#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
using vec = vector<ll>;
using vecp = vector<P>;
using mat = vector<vec>;
using matp = vector<vecp>;
const ll MOD = 998244353;
const ll INF = 1e18;
ll Pow(ll a, ll n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  if (n % 2 == 1) return (a * Pow(a, n - 1));
  ll t = Pow(a, n / 2);
  return (t * t);
}
int main() {
  ll T;
  cin >> T;
  for (long long l = 0; l < (T); ++l) {
    ll N;
    cin >> N;
    vec A(N);
    for (long long i = 0; i < (N); ++i) cin >> A.at(i);
    if (N == 1) {
      if (A.at(0) == 0) {
        cout << "YES" << endl;
      } else {
        cout << "NO" << endl;
      }
      continue;
    }
    ll t = 0;
    for (long long i = 0; i < (N); ++i) {
      for (long long tmp = 0; tmp < (Pow(3, N - 1)); ++tmp) {
        ll k = 0, s = tmp;
        for (long long j = 0; j < (N); ++j) {
          if (j == i) {
            continue;
          }
          if (s % 3 == 1)
            k += A.at(j);
          else if (s % 3 == 2)
            k -= A.at(j);
          s /= 3;
        }
        if (k == A.at(i)) t = 1;
      }
    }
    if (t == 1) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
}

### Prompt

Please formulate 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>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
using vec = vector<ll>;
using vecp = vector<P>;
using mat = vector<vec>;
using matp = vector<vecp>;
const ll MOD = 998244353;
const ll INF = 1e18;
ll Pow(ll a, ll n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  if (n % 2 == 1) return (a * Pow(a, n - 1));
  ll t = Pow(a, n / 2);
  return (t * t);
}
int main() {
  ll T;
  cin >> T;
  for (long long l = 0; l < (T); ++l) {
    ll N;
    cin >> N;
    vec A(N);
    for (long long i = 0; i < (N); ++i) cin >> A.at(i);
    if (N == 1) {
      if (A.at(0) == 0) {
        cout << "YES" << endl;
      } else {
        cout << "NO" << endl;
      }
      continue;
    }
    ll t = 0;
    for (long long i = 0; i < (N); ++i) {
      for (long long tmp = 0; tmp < (Pow(3, N - 1)); ++tmp) {
        ll k = 0, s = tmp;
        for (long long j = 0; j < (N); ++j) {
          if (j == i) {
            continue;
          }
          if (s % 3 == 1)
            k += A.at(j);
          else if (s % 3 == 2)
            k -= A.at(j);
          s /= 3;
        }
        if (k == A.at(i)) t = 1;
      }
    }
    if (t == 1) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
}

```