#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') w = -1;
  for (; isdigit(ch); ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ 48);
  return s * w;
}
inline void print(long long x, int op = 10) {
  if (!x) {
    putchar('0');
    if (op) putchar(op);
    return;
  }
  char F[40];
  long long tmp = x > 0 ? x : -x;
  if (x < 0) putchar('-');
  int cnt = 0;
  while (tmp > 0) {
    F[cnt++] = tmp % 10 + '0';
    tmp /= 10;
  }
  while (cnt > 0) putchar(F[--cnt]);
  if (op) putchar(op);
}
inline long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans *= a;
    b >>= 1;
    a *= a;
  }
  return ans;
}
long long qpow(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) (ans *= a) %= mod;
    b >>= 1;
    (a *= a) %= mod;
  }
  return ans % mod;
}
const int dir[][2] = {{0, 1}, {1, 0},  {0, -1}, {-1, 0},
                      {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long INF64 = 0x3f3f3f3f3f3f3f3f;
const int N = 1e6 + 7;
struct Node {
  long long val;
  int id;
  bool operator<(const Node& opt) const { return val < opt.val; }
};
long long n, m;
set<int> st;
int p[20];
int solve() {
  n = read();
  int cnt = 0;
  for (int i = 1; i <= n; ++i) p[i] = read(), cnt += p[i] == 0;
  st.clear();
  st.insert(0);
  for (int i = 1; i <= n; ++i) {
    set<int> tmp = st;
    for (auto& it : st) {
      int x1 = it - p[i];
      int x2 = it + p[i];
      cnt += x1 == 0;
      cnt += x2 == 0;
      tmp.insert(x1), tmp.insert(x2);
    }
    st = tmp;
  }
  return cnt;
}
int main() {
  int T = read();
  for (int _ = 1; _ <= T; ++_) {
    cout << (solve() ? "YES" : "NO") << "\n";
  }
  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>
using namespace std;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') w = -1;
  for (; isdigit(ch); ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ 48);
  return s * w;
}
inline void print(long long x, int op = 10) {
  if (!x) {
    putchar('0');
    if (op) putchar(op);
    return;
  }
  char F[40];
  long long tmp = x > 0 ? x : -x;
  if (x < 0) putchar('-');
  int cnt = 0;
  while (tmp > 0) {
    F[cnt++] = tmp % 10 + '0';
    tmp /= 10;
  }
  while (cnt > 0) putchar(F[--cnt]);
  if (op) putchar(op);
}
inline long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans *= a;
    b >>= 1;
    a *= a;
  }
  return ans;
}
long long qpow(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) (ans *= a) %= mod;
    b >>= 1;
    (a *= a) %= mod;
  }
  return ans % mod;
}
const int dir[][2] = {{0, 1}, {1, 0},  {0, -1}, {-1, 0},
                      {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long INF64 = 0x3f3f3f3f3f3f3f3f;
const int N = 1e6 + 7;
struct Node {
  long long val;
  int id;
  bool operator<(const Node& opt) const { return val < opt.val; }
};
long long n, m;
set<int> st;
int p[20];
int solve() {
  n = read();
  int cnt = 0;
  for (int i = 1; i <= n; ++i) p[i] = read(), cnt += p[i] == 0;
  st.clear();
  st.insert(0);
  for (int i = 1; i <= n; ++i) {
    set<int> tmp = st;
    for (auto& it : st) {
      int x1 = it - p[i];
      int x2 = it + p[i];
      cnt += x1 == 0;
      cnt += x2 == 0;
      tmp.insert(x1), tmp.insert(x2);
    }
    st = tmp;
  }
  return cnt;
}
int main() {
  int T = read();
  for (int _ = 1; _ <= T; ++_) {
    cout << (solve() ? "YES" : "NO") << "\n";
  }
  return 0;
}

```