#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
inline int read() {
  int ret = 0, sgn = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') sgn = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * sgn;
}
struct ui {
  int fir;
  int sec;
  bool operator<(const ui b) const { return this->fir < b.fir; }
};
const int maxn = 1005;
int sz = 0;
void dfs(map<int, int> &ma, vector<int> &lst, int now, int sum) {
  if (now == sz) {
    return;
  }
  dfs(ma, lst, now + 1, sum);
  sum += lst[now];
  ma[sum] += 1;
  dfs(ma, lst, now + 1, sum);
  sum -= lst[now];
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  int idkvwo;
  cin >> idkvwo;
  while (idkvwo--) {
    int n;
    cin >> n;
    int tag = 0;
    int tmp = 0;
    map<int, int> ma;
    vector<int> lst;
    for (int i = 1; i <= n; ++i) {
      cin >> tmp;
      lst.push_back(abs(tmp));
      if (tmp == 0) {
        tag = 1;
      }
    }
    sz = lst.size();
    if (tag == 1) {
      cout << "YES" << endl;
      continue;
    }
    dfs(ma, lst, 0, 0);
    for (auto i : ma) {
      if (i.second >= 2) {
        tag = 1;
      }
    }
    if (tag)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
}

### Prompt

Please create a solution in cpp 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>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
inline int read() {
  int ret = 0, sgn = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') sgn = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * sgn;
}
struct ui {
  int fir;
  int sec;
  bool operator<(const ui b) const { return this->fir < b.fir; }
};
const int maxn = 1005;
int sz = 0;
void dfs(map<int, int> &ma, vector<int> &lst, int now, int sum) {
  if (now == sz) {
    return;
  }
  dfs(ma, lst, now + 1, sum);
  sum += lst[now];
  ma[sum] += 1;
  dfs(ma, lst, now + 1, sum);
  sum -= lst[now];
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  int idkvwo;
  cin >> idkvwo;
  while (idkvwo--) {
    int n;
    cin >> n;
    int tag = 0;
    int tmp = 0;
    map<int, int> ma;
    vector<int> lst;
    for (int i = 1; i <= n; ++i) {
      cin >> tmp;
      lst.push_back(abs(tmp));
      if (tmp == 0) {
        tag = 1;
      }
    }
    sz = lst.size();
    if (tag == 1) {
      cout << "YES" << endl;
      continue;
    }
    dfs(ma, lst, 0, 0);
    for (auto i : ma) {
      if (i.second >= 2) {
        tag = 1;
      }
    }
    if (tag)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
}

```