#include <bits/stdc++.h>
using namespace std;
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
pair<int, int> operator*(const pair<int, int> &p, const pair<int, int> &x) {
  pair<int, int> np = make_pair(p.first * x.first, p.second * x.second);
  return np;
}
pair<long long, long long> operator%(const pair<long long, long long> &p,
                                     const pair<long long, long long> &x) {
  pair<long long, long long> np =
      make_pair(p.first % x.first, p.second % x.second);
  return np;
}
pair<long long, long long> operator*(const pair<long long, long long> &p,
                                     const pair<int, int> &x) {
  pair<long long, long long> np =
      make_pair(p.first * x.first, p.second * x.second);
  return np;
}
pair<long long, long long> operator*(const pair<long long, long long> &p,
                                     const pair<long long, long long> &x) {
  pair<long long, long long> np =
      make_pair(p.first * x.first, p.second * x.second);
  return np;
}
pair<long long, long long> operator+(const pair<long long, long long> &p,
                                     const pair<long long, long long> &x) {
  pair<long long, long long> np =
      make_pair(p.first + x.first, p.second + x.second);
  return np;
}
pair<long long, long long> operator-(const pair<long long, long long> &p,
                                     const pair<long long, long long> &x) {
  pair<long long, long long> np =
      make_pair(p.first - x.first, p.second - x.second);
  return np;
}
pair<int, int> operator+(const pair<int, int> &p, const pair<int, int> &x) {
  pair<int, int> np = make_pair(p.first + x.first, p.second + x.second);
  return np;
}
pair<int, int> operator-(const pair<int, int> &p, const pair<int, int> &x) {
  pair<int, int> np = make_pair(p.first - x.first, p.second - x.second);
  return np;
}
vector<int> prime;
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long powerm(long long a, long long k) {
  if (!k) {
    return 1;
  }
  long long b = powerm(a, k / 2);
  b = b * b % 998244353;
  if (k % 2) {
    return a * b % 998244353;
  } else {
    return b;
  }
}
long long power(long long a, long long b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  long long m1 = power(a, b / 2);
  if (b % 2) return m1 * m1 * a;
  return m1 * m1;
}
void PFactor(int x) {
  prime.clear();
  for (int i = 2; i <= x / i; i++)
    if (x % i == 0) {
      while (x % i == 0) x /= i, prime.push_back(i);
    }
  if (x > 1) prime.push_back(x);
}
bool isprime(long long a) {
  if (a <= 1) return false;
  if (a == 2 || a == 3) return true;
  if (a % 2 == 0 || a % 3 == 0) return false;
  for (long long i = 5; i * i <= a; i = i + 6) {
    if (a % i == 0 || a % (i + 2) == 0) return false;
  }
  return true;
}
long long read() {
  long long a = 0;
  char b = 1, c;
  do
    if ((c = getchar()) == 45) b = -1;
  while (c < 48 || c > 57);
  do a = (a << 3) + (a << 1) + (c & 15);
  while ((c = getchar()) > 47 && c < 58);
  return a * b;
}
void write(long long x, char c) {
  if (x < 0) putchar(45), x = -x;
  char a[20], s = 0;
  do a[++s] = x % 10 | 48;
  while (x /= 10);
  do putchar(a[s]);
  while (--s);
  putchar(c);
}
const int N = 2e6 + 20;
vector<int> g[N];
int n = 0, m = 0, k = 0;
int a = 0, b = 0, c = 0;
string str;
int d[N] = {0};
int ans[N];
bool flag = false;
void dfs(int deep) {
  if (flag) return;
  if (deep == n + 1) return;
  int x = d[deep];
  for (int i = 0; i <= deep - 1; i++) {
    ans[deep] = ans[i] + x;
    for (int j = 0; j <= deep - 1; j++)
      if (ans[deep] == ans[j]) flag = true;
    dfs(deep + 1);
  }
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = 1; i <= n; i++) d[i] = read();
    flag = false;
    for (int i = 0; i <= n + 10; i++) ans[i] = 0;
    dfs(1);
    if (flag)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
pair<int, int> operator*(const pair<int, int> &p, const pair<int, int> &x) {
  pair<int, int> np = make_pair(p.first * x.first, p.second * x.second);
  return np;
}
pair<long long, long long> operator%(const pair<long long, long long> &p,
                                     const pair<long long, long long> &x) {
  pair<long long, long long> np =
      make_pair(p.first % x.first, p.second % x.second);
  return np;
}
pair<long long, long long> operator*(const pair<long long, long long> &p,
                                     const pair<int, int> &x) {
  pair<long long, long long> np =
      make_pair(p.first * x.first, p.second * x.second);
  return np;
}
pair<long long, long long> operator*(const pair<long long, long long> &p,
                                     const pair<long long, long long> &x) {
  pair<long long, long long> np =
      make_pair(p.first * x.first, p.second * x.second);
  return np;
}
pair<long long, long long> operator+(const pair<long long, long long> &p,
                                     const pair<long long, long long> &x) {
  pair<long long, long long> np =
      make_pair(p.first + x.first, p.second + x.second);
  return np;
}
pair<long long, long long> operator-(const pair<long long, long long> &p,
                                     const pair<long long, long long> &x) {
  pair<long long, long long> np =
      make_pair(p.first - x.first, p.second - x.second);
  return np;
}
pair<int, int> operator+(const pair<int, int> &p, const pair<int, int> &x) {
  pair<int, int> np = make_pair(p.first + x.first, p.second + x.second);
  return np;
}
pair<int, int> operator-(const pair<int, int> &p, const pair<int, int> &x) {
  pair<int, int> np = make_pair(p.first - x.first, p.second - x.second);
  return np;
}
vector<int> prime;
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long powerm(long long a, long long k) {
  if (!k) {
    return 1;
  }
  long long b = powerm(a, k / 2);
  b = b * b % 998244353;
  if (k % 2) {
    return a * b % 998244353;
  } else {
    return b;
  }
}
long long power(long long a, long long b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  long long m1 = power(a, b / 2);
  if (b % 2) return m1 * m1 * a;
  return m1 * m1;
}
void PFactor(int x) {
  prime.clear();
  for (int i = 2; i <= x / i; i++)
    if (x % i == 0) {
      while (x % i == 0) x /= i, prime.push_back(i);
    }
  if (x > 1) prime.push_back(x);
}
bool isprime(long long a) {
  if (a <= 1) return false;
  if (a == 2 || a == 3) return true;
  if (a % 2 == 0 || a % 3 == 0) return false;
  for (long long i = 5; i * i <= a; i = i + 6) {
    if (a % i == 0 || a % (i + 2) == 0) return false;
  }
  return true;
}
long long read() {
  long long a = 0;
  char b = 1, c;
  do
    if ((c = getchar()) == 45) b = -1;
  while (c < 48 || c > 57);
  do a = (a << 3) + (a << 1) + (c & 15);
  while ((c = getchar()) > 47 && c < 58);
  return a * b;
}
void write(long long x, char c) {
  if (x < 0) putchar(45), x = -x;
  char a[20], s = 0;
  do a[++s] = x % 10 | 48;
  while (x /= 10);
  do putchar(a[s]);
  while (--s);
  putchar(c);
}
const int N = 2e6 + 20;
vector<int> g[N];
int n = 0, m = 0, k = 0;
int a = 0, b = 0, c = 0;
string str;
int d[N] = {0};
int ans[N];
bool flag = false;
void dfs(int deep) {
  if (flag) return;
  if (deep == n + 1) return;
  int x = d[deep];
  for (int i = 0; i <= deep - 1; i++) {
    ans[deep] = ans[i] + x;
    for (int j = 0; j <= deep - 1; j++)
      if (ans[deep] == ans[j]) flag = true;
    dfs(deep + 1);
  }
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = 1; i <= n; i++) d[i] = read();
    flag = false;
    for (int i = 0; i <= n + 10; i++) ans[i] = 0;
    dfs(1);
    if (flag)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}

```