#include <bits/stdc++.h>
using namespace std;
const int INF = (-1u) / 2;
const long long int INF2 = (-1ull) / 2;
int a, b, i, d[1011000], j, k, n, m, timer = 0, l, r, x, y;
int c[1011000], cnt = 0, fl = 0, a2, a3 = -1000000, ans = 0;
deque<int> v;
void rec(deque<int> v, int id) {
  if (id == n) {
    if (v.size() == 1 && v[0] == n) {
      printf("YES\n");
      exit(0);
    }
    return;
  }
  deque<int> vv;
  for (int i = id; i < n; i++) {
    if (c[i] == 1)
      v.push_back(1);
    else if (c[i] == 2) {
      printf("NO\n");
      exit(0);
    } else {
      int fl = 0;
      int k = v.size();
      map<int, int> mp;
      for (int m = 1; m < (1 << k); m++) {
        int cnt = 0;
        int ans = 0;
        for (int j = 0; j < k; j++) {
          if ((m & (1 << j))) {
            cnt += v[j];
            ans++;
          }
        }
        if (cnt == c[i] - 1 && ans > 1) {
          ans = 0;
          int msk = 0;
          if (vv.size() > 0) vv.clear();
          for (int j = 0; j < k; j++) {
            if (!(m & (1 << j))) {
              vv.push_back(v[j]);
            } else {
              msk |= (1 << d[v[j]]);
            }
          }
          if (mp[msk]) continue;
          mp[msk] = 1;
          vv.push_back(c[i]);
          rec(vv, i + 1);
          fl = 1;
        }
      }
      if (fl == 0) {
        return;
      }
    }
  }
  if (v.size() == 1 && v[0] == n) {
    printf("YES\n");
    exit(0);
  }
}
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &c[i]);
  }
  sort(c, c + n);
  for (i = 0; i < n; i++) {
    d[c[i]] = i;
  }
  rec(v, 0);
  printf("NO\n");
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food. 

Iahub asks Iahubina: can you build a rooted tree, such that

  * each internal node (a node with at least one son) has at least two sons; 
  * node i has ci nodes in its subtree? 



Iahubina has to guess the tree. Being a smart girl, she realized that it's possible no tree can follow Iahub's restrictions. In this way, Iahub will eat all the food. You need to help Iahubina: determine if there's at least one tree following Iahub's restrictions. The required tree must contain n nodes.

Input

The first line of the input contains integer n (1 ≤ n ≤ 24). Next line contains n positive integers: the i-th number represents ci (1 ≤ ci ≤ n).

Output

Output on the first line "YES" (without quotes) if there exist at least one tree following Iahub's restrictions, otherwise output "NO" (without quotes). 

Examples

Input

4
1 1 1 4


Output

YES

Input

5
1 1 5 2 1


Output

NO

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = (-1u) / 2;
const long long int INF2 = (-1ull) / 2;
int a, b, i, d[1011000], j, k, n, m, timer = 0, l, r, x, y;
int c[1011000], cnt = 0, fl = 0, a2, a3 = -1000000, ans = 0;
deque<int> v;
void rec(deque<int> v, int id) {
  if (id == n) {
    if (v.size() == 1 && v[0] == n) {
      printf("YES\n");
      exit(0);
    }
    return;
  }
  deque<int> vv;
  for (int i = id; i < n; i++) {
    if (c[i] == 1)
      v.push_back(1);
    else if (c[i] == 2) {
      printf("NO\n");
      exit(0);
    } else {
      int fl = 0;
      int k = v.size();
      map<int, int> mp;
      for (int m = 1; m < (1 << k); m++) {
        int cnt = 0;
        int ans = 0;
        for (int j = 0; j < k; j++) {
          if ((m & (1 << j))) {
            cnt += v[j];
            ans++;
          }
        }
        if (cnt == c[i] - 1 && ans > 1) {
          ans = 0;
          int msk = 0;
          if (vv.size() > 0) vv.clear();
          for (int j = 0; j < k; j++) {
            if (!(m & (1 << j))) {
              vv.push_back(v[j]);
            } else {
              msk |= (1 << d[v[j]]);
            }
          }
          if (mp[msk]) continue;
          mp[msk] = 1;
          vv.push_back(c[i]);
          rec(vv, i + 1);
          fl = 1;
        }
      }
      if (fl == 0) {
        return;
      }
    }
  }
  if (v.size() == 1 && v[0] == n) {
    printf("YES\n");
    exit(0);
  }
}
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &c[i]);
  }
  sort(c, c + n);
  for (i = 0; i < n; i++) {
    d[c[i]] = i;
  }
  rec(v, 0);
  printf("NO\n");
  return 0;
}

```