Codeforces1656C题 Make Equal With Mod

Codeforces1656C题 Make Equal With Mod

题目

C. Make Equal With Mod time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output

You are given an array of n non-negative integers a1,a2,…,an. You can make the following operation: choose an integer x≥2x≥2 and replace each number of the array by the remainder when dividing that number by xx, that is, for all 1≤i≤n set aiai to aimodxaimodx.

Determine if it is possible to make all the elements of the array equal by applying the operation zero or more times.

Input

The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤10^4) — the number of test cases. Description of the test cases follows.

The first line of each test case contains an integer nn (1≤n≤10^5) — the length of the array.

The second line of each test case contains nn integers a1,a2,…,an (0≤ai≤10^9) where aiai is the ii-th element of the array.

The sum of n for all test cases is at most 2⋅10^5.

Output

For each test case, print a line with YES if you can make all elements of the list equal by applying the operation. Otherwise, print NO.

You may print each letter in any case (for example, "YES", "Yes", "yes", "yEs" will all be recognized as a positive answer).

Example

input

Copy

4
4
2 5 6 8
3
1 1 1
5
4 1 7 0 8
4
5 9 17 5

output

Copy

YES
YES
NO
YES

Note

In the first test case, one can apply the operation with x=3 to obtain the array [2,2,0,2][2,2,0,2], and then apply the operation with x=2 to obtain [0,0,0,0][0,0,0,0].

In the second test case, all numbers are already equal.

In the fourth test case, applying the operation with x=4 results in the array [1,1,1,1][1,1,1,1].

思路

1. 当数组中一开始没有1，对于数组中任意数n,我们利用n%n=0可以从最大项开始将每一项变为0
2. 当数组中一开始有1，但不存在两个数仅相差1的情况下，对于数组中任意数n，利用n%(n-1)=1可以从最大项开始将每一项变为1

代码

#include<iostream>
#include<algorithm>

using namespace std;

typedef long long ll;

int T,n,k;
ll a[200005];

int main() {
	cin>>T;
	while(T--) {
		cin>>n;
		int f1=0,f2=0;
		for(int i=0; i<n; i++) {
			cin>>a[i];
			if(a[i]==1) f1=1;
		}
		sort(a,a+n);
		for(int i=1; i<n; i++)
			if(a[i]-a[i-1]==1) f2=1;
		if(!f1||!f2)
			puts("Yes");
		else 
			puts("No");
	}
}