ModularEquation.cpp

/*
D. Moderate Modular Mode
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
YouKn0wWho has two even integers x and y. Help him to find an integer n such that 1≤n≤2⋅1018 and nmodx=ymodn. Here, amodb denotes the remainder of a after division by b. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints.

Input
The first line contains a single integer t (1≤t≤105)  — the number of test cases.

The first and only line of each test case contains two integers x and y (2≤x,y≤109, both are even).

Output
For each test case, print a single integer n (1≤n≤2⋅1018) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints.

Example
inputCopy
4
4 8
4 2
420 420
69420 42068
outputCopy
4
10
420
9969128
Note
In the first test case, 4mod4=8mod4=0.

In the second test case, 10mod4=2mod10=2.

In the third test case, 420mod420=420mod420=0.


*/
#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <chrono>
#include <complex>
using namespace std;
#define ll long long
#define ld long double
#define ui unsigned int
#define ull unsigned ll
#define mp                   make_pair
#define eb                   emplace_back
#define pb                   push_back
#define pf                   push_front
#define popb                 pop_back
#define popf                 pop_front
#define hashmap              unordered_map
#define hashset              unordered_set
#define lb                   lower_bound
#define ub                   upper_bound
#define all(a)               (a).begin(), (a).end()
#define rall(a)              (a).rbegin(), (a).rend()
#define ff                   first
#define ss                   second
#define foi(n) for(ll i=0;i<n;i++)
#define foj(n) for(ll j=0;j<n;j++)
#define fok(n) for(ll k=0;k<n;k++)
#define forr(i,a,b) for(ll i=a;i<b;i++)
#define forrr(i,b,a) for(ll i=b;i>=a;i--)
#define forrrr(i,a,b,k) for(ll i=a;i<b;i=i+k)
#define graph          vector<vector<ll>>
#define sz(v) v.size()
typedef pair<int, int> pii;
typedef pair<ll, ll>   pll;
typedef vector<int>         vi;
typedef vector<ll>          vll;
typedef vector<string>      vs;
typedef vector<double>      vd;
typedef vector<pii>         vpii;
typedef vector<pll>         vpll;
typedef pair< ll, pll>      plll;
typedef queue<ll>           qll;
typedef vector<plll>      vplll;
typedef  vector<set<ll>>   vsll;
typedef  vector<char>	         vc;
typedef  vector<bool>            vb;
typedef  map<string, int>         msi;
typedef  map<int, int>	         mii;
typedef  map<ll, ll>             mll;
typedef  map<ll, vll>             mvll;
typedef  map<vll, ll>             mvlll;
typedef  map<char, ll>           mcl;
typedef map<pair<ll, ll>, ll>  mplll;
typedef  unordered_map<char, ll>           umcl;
typedef  unordered_map< ll, char>           umlc;
typedef  unordered_map< ll, ld>           umld;
typedef  map<int, string>	     mis;
typedef  pair<string, int>       psi;
typedef  pair<string, string>    pss;
typedef priority_queue <ll> pq;
typedef priority_queue<pii, vector<pii>, greater<pii> > pqq;
typedef priority_queue<ll, vector<ll>, greater<ll>> prq;
const ll MOD = 1000000007;
const ll modx = 998244353;
ld PI = 3.1415926535897;
const ll N = 200010;
void solve();
int main()
{
	ios_base::sync_with_stdio(false); cin.tie(NULL);

#ifndef ONLINE_JUDGE
	freopen("input1.txt", "r", stdin);
	freopen("error1.txt", "w", stderr);
	freopen("output1.txt", "w", stdout);
#endif
	ll t ; cin >> t;
	while (t--)
	{
		solve();
		cout << "\n";
	}

	cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
	return 0;
}
ll ceils(ll x, ll y) {
	return x / y + ((x % y) != 0);
}
ll gcd(ll a, ll b) {
	if (b == 0)
		return a;
	else
		return gcd(b, a % b);
}


ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;

}
ll powmod(ll x, ll y) {
	ll res = 1;
	for (ll i = 0; i < y; i++) {
		res = res * x % MOD;
	}
	return res;
}
bool isPrime(ll n)
{
	if (n <= 1)
		return false;
	if (n <= 3)
		return true;
	if (n % 2 == 0 || n % 3 == 0)
		return false;

	for (ll i = 5; i * i <= n; i = i + 6)
		if (n % i == 0 || n % (i + 2) == 0)
			return false;

	return true;
}
bool COMP(pll l, pll r) {
	return l.ss < r.ss;
}
ll kadanesAlgo(vll a)
{
	ll n = a.size();

	ll currMax = 0;
	ll mx = INT_MIN;

	foi(n)
	{
		currMax += a[i];

		if (currMax <= a[i])
		{
			currMax = a[i];
		}

		mx = max(currMax, mx);
	}
	return mx;
}
ll ask(ll x, ll y, ll n) {
	cout << "?" << " ";
	foi(n - 1) {
		cout << x << " ";
	}
	cout << y << endl;
	ll s;
	cin >> s;
	return s;
}
ll binpow(ll a, ll b, ll m) {
	a %= m;
	ll res = 1;
	while (b > 0) {
		if (b & 1)
			res = res * a % m;
		a = a * a % m;
		b >>= 1;
	}
	return res;
}
bool is_sorted(vector<ll>a) {
	ll x = 0;
	for (ll i = 0; i < a.size() - 1; i++) {
		if (a[i] < a[i + 1])
			x++;
	}
	return x == a.size() - 1;
}
bool check(string a, string b) {
	ll n = a.size(), m = b.size();
	if (a < b) {
		return true;
	}
	if (a > b) {
		return false;
	}
	foi(a.size()) {
		if (a[i] > b[i]) {
			return false;
		}
		else if (a[i] < b[i]) {
			return true;
		}
	}
	return false;
}
bool isSubSequence(string a, string b, ll m, ll n) {
	ll j = 0;
	for (ll i = 0; i < n and j < m; i++)
		if (a[j] == b[i])
			j++;
	return (j == m);
}
ll power(ll n, ll x) {
	ll ans = 1;
	foi(n) {
		ans *= x;
	}
	return ans;
}
void solve() {
	ll n;
	cin >> n;
	vll a(n, 0);
	foi(n) {
		cin >> a[i];
	}
	while (a.size()) {
		ll idx = a.size() - 1;
		while (idx > -1 and a[idx] % (idx + 2) == 0) {
			idx--;
		}
		if (idx < 0) {
			cout << "No";
			return;
		}
		a.erase(a.begin() + idx);
	}
	cout << "Yes";
}