#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
template <class T>
void sort_unique(vector<T> &v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
int N;
int X[1 << 10];
int A[10];
void MAIN() {
  scanf("%d", &N);
  for (int i = 0, i_len = (N); i < i_len; ++i) {
    int x;
    scanf("%d", &x);
    A[i] = abs(x);
  }
  bool yes = false;
  for (int s = 1; s < (1 << N); s++) {
    int k = __builtin_ctz(s);
    X[s] = X[s ^ (1 << k)] + A[k];
  }
  sort(X, X + (1 << N));
  for (int i = 0, i_len = ((1 << N) - 1); i < i_len; ++i) {
    if (X[i] == X[i + 1]) yes = true;
  }
  puts(yes ? "YES" : "NO");
}
int main() {
  int TC = 1;
  scanf("%d", &TC);
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

You are given a sequence of n integers a_1,   a_2,   ...,   a_n.

Does there exist a sequence of n integers b_1,   b_2,   ...,   b_n such that the following property holds?

  * For each 1 ≤ i ≤ n, there exist two (not necessarily distinct) indices j and k (1 ≤ j,   k ≤ n) such that a_i = b_j - b_k. 

Input

The first line contains a single integer t (1 ≤ t ≤ 20) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 10).

The second line of each test case contains the n integers a_1,   ...,   a_n (-10^5 ≤ a_i ≤ 10^5).

Output

For each test case, output a line containing YES if a sequence b_1,   ...,   b_n satisfying the required property exists, and NO otherwise.

Example

Input


5
5
4 -7 -1 5 10
1
0
3
1 10 100
4
-3 2 10 2
9
25 -171 250 174 152 242 100 -205 -258


Output


YES
YES
NO
YES
YES

Note

In the first test case, the sequence b = [-9,   2,   1,   3,   -2] satisfies the property. Indeed, the following holds: 

  * a_1 = 4 = 2 - (-2) = b_2 - b_5; 
  * a_2 = -7 = -9 - (-2) = b_1 - b_5; 
  * a_3 = -1 = 1 - 2 = b_3 - b_2; 
  * a_4 = 5 = 3 - (-2) = b_4 - b_5; 
  * a_5 = 10 = 1 - (-9) = b_3 - b_1. 



In the second test case, it is sufficient to choose b = [0], since a_1 = 0 = 0 - 0 = b_1 - b_1.

In the third test case, it is possible to show that no sequence b of length 3 satisfies the property.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
template <class T>
void sort_unique(vector<T> &v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
int N;
int X[1 << 10];
int A[10];
void MAIN() {
  scanf("%d", &N);
  for (int i = 0, i_len = (N); i < i_len; ++i) {
    int x;
    scanf("%d", &x);
    A[i] = abs(x);
  }
  bool yes = false;
  for (int s = 1; s < (1 << N); s++) {
    int k = __builtin_ctz(s);
    X[s] = X[s ^ (1 << k)] + A[k];
  }
  sort(X, X + (1 << N));
  for (int i = 0, i_len = ((1 << N) - 1); i < i_len; ++i) {
    if (X[i] == X[i + 1]) yes = true;
  }
  puts(yes ? "YES" : "NO");
}
int main() {
  int TC = 1;
  scanf("%d", &TC);
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}

```