#include <bits/stdc++.h>
using namespace std;
int sz;
int a[14];
int dp[14][(1LL << 14) + 10][15];
int solve(int pos, int mask, int leaf) {
  int &ret = dp[pos][mask][leaf];
  if (ret != -1) return ret;
  if (pos == sz) {
    if (mask == 0 && leaf == 0) return 1;
    return 0;
  }
  if (mask & (1LL << pos)) return ret = 0;
  ret = 0;
  for (int nmask = mask;; nmask = (nmask - 1) & mask) {
    int sum = 0;
    int cnt = 0;
    for (int i = pos + 1; i < sz; i++) {
      if (nmask & (1LL << i)) sum += a[i], cnt++;
    }
    int rem = a[pos] - 1 - sum;
    if (leaf >= rem && rem + cnt >= 2) {
      ret |= solve(pos + 1, nmask ^ mask, leaf - rem);
    }
    if (nmask == 0) break;
  }
  return ret;
}
int main() {
  int cs, ts, i, j, k, x, y, z, q, m, n, leaf = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x == 1) {
      leaf++;
    } else {
      a[sz++] = x;
    }
  }
  if (n == 1 && leaf == 1) {
    printf("YES\n");
    return 0;
  }
  if (leaf < (n + 1) / 2) {
    printf("NO\n");
    return 0;
  }
  sort(a, a + sz, greater<int>());
  memset(dp, -1, sizeof(dp));
  int ret = solve(0, (((1LL << sz) - 1) | 1) ^ 1, leaf);
  if (ret) {
    printf("YES\n");
  } else {
    printf("NO\n");
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int sz;
int a[14];
int dp[14][(1LL << 14) + 10][15];
int solve(int pos, int mask, int leaf) {
  int &ret = dp[pos][mask][leaf];
  if (ret != -1) return ret;
  if (pos == sz) {
    if (mask == 0 && leaf == 0) return 1;
    return 0;
  }
  if (mask & (1LL << pos)) return ret = 0;
  ret = 0;
  for (int nmask = mask;; nmask = (nmask - 1) & mask) {
    int sum = 0;
    int cnt = 0;
    for (int i = pos + 1; i < sz; i++) {
      if (nmask & (1LL << i)) sum += a[i], cnt++;
    }
    int rem = a[pos] - 1 - sum;
    if (leaf >= rem && rem + cnt >= 2) {
      ret |= solve(pos + 1, nmask ^ mask, leaf - rem);
    }
    if (nmask == 0) break;
  }
  return ret;
}
int main() {
  int cs, ts, i, j, k, x, y, z, q, m, n, leaf = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x == 1) {
      leaf++;
    } else {
      a[sz++] = x;
    }
  }
  if (n == 1 && leaf == 1) {
    printf("YES\n");
    return 0;
  }
  if (leaf < (n + 1) / 2) {
    printf("NO\n");
    return 0;
  }
  sort(a, a + sz, greater<int>());
  memset(dp, -1, sizeof(dp));
  int ret = solve(0, (((1LL << sz) - 1) | 1) ^ 1, leaf);
  if (ret) {
    printf("YES\n");
  } else {
    printf("NO\n");
  }
  return 0;
}

```