#include <bits/stdc++.h>
using namespace std;
const long long mod1 = (long long)1e9 + 7;
const long long mod2 = 998244353;
const long long MAXN = (long long)1e6 + 7;
const long double pi = acos(-1);
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long pow1(long long p, long long d, long long P) {
  return d ? (pow1(p * p % P, d >> 1, P) * (d & 1 ? p % P : 1)) % P : 1;
}
vector<long long> prime;
bool is_composite[MAXN];
void sieve(long long n) {
  std::fill(is_composite, is_composite + n, false);
  for (long long i = 2; i < n; ++i) {
    if (!is_composite[i]) prime.push_back(i);
    for (auto &p : prime) {
      if (i * p >= n) break;
      is_composite[i * p] = true;
      if (i % p == 0) break;
    }
  }
}
pair<long long, long long> ExtendedEuclid(long long a, long long b) {
  long long s1, s2, t1, t2;
  s1 = t2 = 1;
  s2 = t1 = 0;
  while (a % b != 0) {
    long long q = a / b;
    long long s3 = s2;
    s2 = s1 - q * s2;
    s1 = s3;
    s3 = t2;
    t2 = t1 - q * t2;
    t1 = s3;
    s3 = b;
    b = a % b;
    a = s3;
  }
  return {s2, t2};
}
void doit() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    a[i] = abs(a[i]);
  }
  if (*min_element(a.begin(), a.end()) == 0) {
    cout << "YES";
    return;
  }
  set<long long> sums;
  for (long long p = 0; p < (1 << n); ++p) {
    long long sum = 0;
    for (long long i = 0; i < n; ++i) {
      if (p & (1 << i)) sum += a[i];
    }
    if (sums.find(sum) != sums.end()) {
      cout << "YES";
      return;
    }
    sums.insert(sum);
  }
  cout << "NO";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  for (cin >> t; t--; cout << '\n') doit();
  return 0;
}

### Prompt

Create a solution in Cpp for 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 long long mod1 = (long long)1e9 + 7;
const long long mod2 = 998244353;
const long long MAXN = (long long)1e6 + 7;
const long double pi = acos(-1);
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long pow1(long long p, long long d, long long P) {
  return d ? (pow1(p * p % P, d >> 1, P) * (d & 1 ? p % P : 1)) % P : 1;
}
vector<long long> prime;
bool is_composite[MAXN];
void sieve(long long n) {
  std::fill(is_composite, is_composite + n, false);
  for (long long i = 2; i < n; ++i) {
    if (!is_composite[i]) prime.push_back(i);
    for (auto &p : prime) {
      if (i * p >= n) break;
      is_composite[i * p] = true;
      if (i % p == 0) break;
    }
  }
}
pair<long long, long long> ExtendedEuclid(long long a, long long b) {
  long long s1, s2, t1, t2;
  s1 = t2 = 1;
  s2 = t1 = 0;
  while (a % b != 0) {
    long long q = a / b;
    long long s3 = s2;
    s2 = s1 - q * s2;
    s1 = s3;
    s3 = t2;
    t2 = t1 - q * t2;
    t1 = s3;
    s3 = b;
    b = a % b;
    a = s3;
  }
  return {s2, t2};
}
void doit() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    a[i] = abs(a[i]);
  }
  if (*min_element(a.begin(), a.end()) == 0) {
    cout << "YES";
    return;
  }
  set<long long> sums;
  for (long long p = 0; p < (1 << n); ++p) {
    long long sum = 0;
    for (long long i = 0; i < n; ++i) {
      if (p & (1 << i)) sum += a[i];
    }
    if (sums.find(sum) != sums.end()) {
      cout << "YES";
      return;
    }
    sums.insert(sum);
  }
  cout << "NO";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  for (cin >> t; t--; cout << '\n') doit();
  return 0;
}

```