A Array with nums

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array aa consisting of nn integers.

In one move, you can choose two indices 1≤i,j≤n1≤i,j≤n such that i≠ji≠j and set ai:=ajai:=aj. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose ii and jj and replace aiai with ajaj).

Your task is to say if it is possible to obtain an array with an odd (not divisible by 22) sum of elements.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤20001≤t≤2000) — the number of test cases.

The next 2t2t lines describe test cases. The first line of the test case contains one integer nn (1≤n≤20001≤n≤2000) — the number of elements in aa. The second line of the test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤20001≤ai≤2000), where aiai is the ii-th element of aa.

It is guaranteed that the sum of nn over all test cases does not exceed 20002000 (∑n≤2000∑n≤2000).

Output

For each test case, print the answer on it — "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.

Example

input

5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1

output

YES
NO
YES
NO
NO

 

在看到這題目的時候有點兒懵逼，沒看懂英文什麼意思，仔細研究了以下發現題意大致如下：

假設有一個數組a,裏面的數值可以被賦值，例如：a[i] = a[j];找出是否有可能使數組中的和爲奇數，若可以則輸出yes，否則輸出no;

有這樣一種方法：

假設數組包含n個元素，若n個元素中含有偶數個奇數，其餘均爲奇數，根據題意可知，偶數可以被替換爲奇數（不一定被全部替換）

可以查找數組中目前有多少個奇數:(有以下兩個極端情況)

若奇數的個數爲n，則說明該數組中每個數字都是奇數。若n爲偶數，則不可能。否則就有可能。

若奇數個數爲0個，則一定不能使數組中的和爲奇數。

代碼如下：

import java.util.Scanner;
 
public class Arraywithodd {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		for(int i=1;i <= n;i++) {
			int t = sc.nextInt();
			int data[] = new int[t];
			int sum = 0;//記錄是否是偶數個奇數
			for(int j = 0;j<t;j++) {
				data[j]=sc.nextInt();
				if(data[j] % 2 !=0)
				{
					sum++;
				}
			}
			if((t % 2 == 0 && sum == t) || sum == 0) {
				System.out.println("NO");
			}else {
				System.out.println("YES");
			}
		}
	}
}