#include <bits/stdc++.h>
template <class T>
inline void read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    f |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  x = f ? -x : x;
  return;
}
std::vector<int> s, t;
int a[20], b[20];
int T, n;
bool flag;
void dfs(int x) {
  if (flag) return;
  if (x > n) {
    if (s.empty() && t.empty()) return;
    int s1 = 0, s2 = 0;
    for (auto i : s) s1 += i;
    for (auto i : t) s2 += i;
    if (s1 == s2) flag = true;
    return;
  }
  dfs(x + 1);
  s.push_back(a[x]);
  dfs(x + 1);
  s.pop_back();
  t.push_back(a[x]);
  dfs(x + 1);
  t.pop_back();
}
int main() {
  read(T);
  while (T--) {
    flag = false;
    s.clear(), t.clear();
    read(n);
    for (int i = 1; i <= n; ++i) read(a[i]);
    dfs(1);
    puts(flag ? "YES" : "NO");
  }
  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>
template <class T>
inline void read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    f |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  x = f ? -x : x;
  return;
}
std::vector<int> s, t;
int a[20], b[20];
int T, n;
bool flag;
void dfs(int x) {
  if (flag) return;
  if (x > n) {
    if (s.empty() && t.empty()) return;
    int s1 = 0, s2 = 0;
    for (auto i : s) s1 += i;
    for (auto i : t) s2 += i;
    if (s1 == s2) flag = true;
    return;
  }
  dfs(x + 1);
  s.push_back(a[x]);
  dfs(x + 1);
  s.pop_back();
  t.push_back(a[x]);
  dfs(x + 1);
  t.pop_back();
}
int main() {
  read(T);
  while (T--) {
    flag = false;
    s.clear(), t.clear();
    read(n);
    for (int i = 1; i <= n; ++i) read(a[i]);
    dfs(1);
    puts(flag ? "YES" : "NO");
  }
  return 0;
}

```