#include <bits/stdc++.h>
using namespace std;
int getint() {
  unsigned int c;
  int x = 0;
  while (((c = getchar()) - '0') >= 10) {
    if (c == '-') return -getint();
    if (!~c) exit(0);
  }
  do {
    x = (x << 3) + (x << 1) + (c - '0');
  } while (((c = getchar()) - '0') < 10);
  return x;
}
int n, as[33], have[33];
int dfs(int u, int need, int lastCheck, int depth) {
  if (u == n) return 1;
  if (need == 1) {
    if (depth != 1) {
      have[as[u]]++;
      return dfs(u + 1, as[u + 1], n, 0);
    }
  } else {
    for (int m = min(need - 1, lastCheck); m >= 1; m--)
      if (have[m]) {
        have[m]--;
        if (dfs(u, need - m, m, depth + 1)) return 1;
        have[m]++;
      }
  }
  return 0;
}
int solve() {
  if (as[n - 1] != n) {
    return 0;
  }
  if (count(as, as + n, 2)) {
    return 0;
  }
  return dfs(0, as[0], n, 0);
}
int main() {
  int i, j, tcc, tc = 1 << 28;
  for (tcc = 0; tcc < tc; tcc++) {
    n = getint();
    memset(have, 0, sizeof(have));
    for (i = 0; i < n; i++) as[i] = getint();
    sort(as, as + n);
    if (solve()) {
      puts("YES");
    } else {
      puts("NO");
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
int getint() {
  unsigned int c;
  int x = 0;
  while (((c = getchar()) - '0') >= 10) {
    if (c == '-') return -getint();
    if (!~c) exit(0);
  }
  do {
    x = (x << 3) + (x << 1) + (c - '0');
  } while (((c = getchar()) - '0') < 10);
  return x;
}
int n, as[33], have[33];
int dfs(int u, int need, int lastCheck, int depth) {
  if (u == n) return 1;
  if (need == 1) {
    if (depth != 1) {
      have[as[u]]++;
      return dfs(u + 1, as[u + 1], n, 0);
    }
  } else {
    for (int m = min(need - 1, lastCheck); m >= 1; m--)
      if (have[m]) {
        have[m]--;
        if (dfs(u, need - m, m, depth + 1)) return 1;
        have[m]++;
      }
  }
  return 0;
}
int solve() {
  if (as[n - 1] != n) {
    return 0;
  }
  if (count(as, as + n, 2)) {
    return 0;
  }
  return dfs(0, as[0], n, 0);
}
int main() {
  int i, j, tcc, tc = 1 << 28;
  for (tcc = 0; tcc < tc; tcc++) {
    n = getint();
    memset(have, 0, sizeof(have));
    for (i = 0; i < n; i++) as[i] = getint();
    sort(as, as + n);
    if (solve()) {
      puts("YES");
    } else {
      puts("NO");
    }
  }
  return 0;
}

```