A. Redstone? - 技术栈

time limit per test

1 second

memory limit per test

256 megabytes

Steve stumbled upon a collection of n gears, where gear i has ai teeth, and he wants to arrange them into a row.

After he arranges them, Steve will spin the leftmost gear at a speed of 1 revolution per second. For each of the other gears, let x be the number of teeth it has, y be the number of teeth of the gear to its left, and z be the speed the gear to its left spins at. Then, its speed will be yx⋅z revolutions per second.

Steve considers the contraption satisfactory if the rightmost gear spins at a speed of 1 revolution per second. Determine whether Steve can rearrange the gears into a satisfactory contraption.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤1000). The description of the test cases follows.

The first line of each test case contains a single integer n (2≤n≤100) --- the number of gears Steve has.

The second line of each test case contains n integers a1,a2,...,an (2≤ai≤100) --- the number of teeth of each gear.

Output

For each test case, print "YES" if Steve can rearrange the gears in a satisfactory way, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Example

Input

Copy

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5 5

6 3 6 9

2 3

30 10 12 10 10 9 18

2 4 8 16 32

Output

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YES
YES
NO
YES
NO

Note

In the first test case, the second gear will always spin at speed 55⋅1=1 revolution per second, so any arrangement is satisfactory.

In the second test case, one possible arrangement is [6,3,9,6]. Then:

The second gear spins at speed 63⋅1=2 revolutions per second.
The third gear spins at speed 39⋅2=23 revolutions per second.
The fourth spins at speed 96⋅23=1 revolution per second.

Since the rightmost gear spins at a speed of 1 revolution per second, the arrangement is satisfactory.

In the third test case, neither of the possible arrangements [2,3] and [3,2] are satisfactory.

解题说明：水题，找规律能发现只需要数列中有两个一样的数字即可。

cpp 复制代码

#include <stdio.h>

int main() 
{
    int t;
    scanf("%d", &t);
    while (t--) 
    {
        int n;
        scanf("%d", &n);
        int a[105] = { 0 };
        int ok = 0;
        for (int i = 0; i < n; i++) 
        {
            int x;
            scanf("%d", &x);
            a[x]++;
            if (a[x] >= 2)
            {
                ok = 1;
            }
        }
        printf(ok ? "YES\n" : "NO\n");
    }
    return 0;
}