Codeforces Round #661 (Div. 3) A. Remove Smallest

You are given the array aa a consisting of n positive (greater than zero) integers.

In one move, you can choose two indices i and j (i≠j) such that the absolute difference between \(a_i\)and \(a_j\)is no more than one (|ai−aj|≤1) and remove the smallest of these two elements. If two elements are equal, you can remove any of them (but exactly one).

Your task is to find if it is possible to obtain the array consisting of only one element using several (possibly, zero) such moves or not.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1≤t≤1000) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1≤n≤50) — the length of aa a . The second line of the test case contains n integers \(a_1,a_2...a_n\) (1≤\(a_i\)≤100), where \(a_i\) is the i -th element of a .

Output

For each test case, print the answer: "YES" if it is possible to obtain the array consisting of only one element using several (possibly, zero) moves described in the problem statement, or "NO" otherwise.

Example

Input

Copy

5
3
1 2 2
4
5 5 5 5
3
1 2 4
4
1 3 4 4
1
100

Output

Copy

YES
YES
NO
NO
YES

水，排序后扫一遍判断相邻的数是否相差小于等于1即可。

#include 
using namespace std;
int a[55];
int main()
{
	int t;
	cin >> t;
	while(t--)
	{
		int n;
		cin >> n;
		for(int i = 1; i <= n; i++)
		{
			cin >> a[i];
		}
		sort(a + 1, a + n +1);
		bool flag = 1;
		for(int i = 2; i <= n; i++)
		{
			if(a[i] > a[i - 1] + 1)
			{
				cout << "NO" << endl;
				flag = 0;
				break;
			}
		} 
		if(flag) cout << "YES" << endl;
	}
	
	return 0;
}