#include <bits/stdc++.h>
int n, a[30], cnt[4100], f[4100][30][2];
int check(int x, int y, int z) {
  if (f[x][y][z] != -1) return (f[x][y][z]);
  int &ret = f[x][y][z];
  if (x == 0) return (y == 0 && z == 0);
  for (int i = 1; i <= n; i++) {
    if (!(x >> i - 1 & 1)) continue;
    int state = x ^ 1 << i - 1;
    for (int j = state;; j = state & j - 1) {
      for (int k = 0; k <= y; k++) {
        int son = cnt[j] + k;
        if (son < a[i] && a[i] - son - 1 <= y - k) {
          int p = a[i] - son - 1;
          if (z == 1 && state == j || !check(state ^ j, y - k - p, 0)) continue;
          if (j == 0 && k == 0 && p >= 2 || j > 0 && check(j, k, p == 0))
            return (ret = true);
        }
      }
      if (j == 0) break;
    }
  }
  return (ret = false);
}
int main() {
  scanf("%d", &n);
  int leaf = 0, m = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      leaf++;
    else
      a[++m] = a[i];
  }
  if (n == 1)
    printf("%s\n", leaf ? "YES" : "NO");
  else {
    n -= leaf;
    if (n >= leaf)
      printf("NO\n");
    else {
      memset(f, -1, sizeof(f));
      for (int i = 0; i < 1 << n; i++) cnt[i] = __builtin_popcount(i);
      bool flag = false;
      for (int i = 1; i <= n; i++)
        if (a[i] == n + leaf) flag = true;
      printf("%s\n", flag && check((1 << n) - 1, leaf, 0) ? "YES" : "NO");
    }
  }
  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>
int n, a[30], cnt[4100], f[4100][30][2];
int check(int x, int y, int z) {
  if (f[x][y][z] != -1) return (f[x][y][z]);
  int &ret = f[x][y][z];
  if (x == 0) return (y == 0 && z == 0);
  for (int i = 1; i <= n; i++) {
    if (!(x >> i - 1 & 1)) continue;
    int state = x ^ 1 << i - 1;
    for (int j = state;; j = state & j - 1) {
      for (int k = 0; k <= y; k++) {
        int son = cnt[j] + k;
        if (son < a[i] && a[i] - son - 1 <= y - k) {
          int p = a[i] - son - 1;
          if (z == 1 && state == j || !check(state ^ j, y - k - p, 0)) continue;
          if (j == 0 && k == 0 && p >= 2 || j > 0 && check(j, k, p == 0))
            return (ret = true);
        }
      }
      if (j == 0) break;
    }
  }
  return (ret = false);
}
int main() {
  scanf("%d", &n);
  int leaf = 0, m = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      leaf++;
    else
      a[++m] = a[i];
  }
  if (n == 1)
    printf("%s\n", leaf ? "YES" : "NO");
  else {
    n -= leaf;
    if (n >= leaf)
      printf("NO\n");
    else {
      memset(f, -1, sizeof(f));
      for (int i = 0; i < 1 << n; i++) cnt[i] = __builtin_popcount(i);
      bool flag = false;
      for (int i = 1; i <= n; i++)
        if (a[i] == n + leaf) flag = true;
      printf("%s\n", flag && check((1 << n) - 1, leaf, 0) ? "YES" : "NO");
    }
  }
  return (0);
}

```