#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 26;
int n, m;
int a[N];
bool dp[N][(1 << N / 2)][N];
bool solv() {
  scanf("%d", &n);
  int qq = 0;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    if (x == 1)
      ++qq;
    else
      a[m++] = x;
  }
  n = m;
  m = qq;
  if (m < n) return false;
  if (m == 1 && n == 0) return true;
  sort(a, a + n);
  dp[0][0][m] = true;
  for (int i = 0; i < n; ++i) {
    for (int x = 0; x < (1 << i); ++x) {
      for (int q = 0; q <= m; ++q) {
        if (!dp[i][x][q]) continue;
        for (int y = x;; y = ((y - 1) & x)) {
          int s = 0;
          int z = 0;
          for (int j = 0; j < n; ++j) {
            if ((y & (1 << j))) {
              ++z;
              s += a[j];
            }
          }
          if (s > (a[i] - 1)) {
            if (y == 0) break;
            continue;
          }
          if (s + q < (a[i] - 1)) {
            if (y == 0) break;
            continue;
          }
          z += ((a[i] - 1) - s);
          if (z >= 2)
            dp[i + 1][((x ^ y) | (1 << i))][q - ((a[i] - 1) - s)] = true;
          if (y == 0) break;
        }
      }
    }
  }
  return dp[n][(1 << (n - 1))][0];
}
int main() {
  if (solv())
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 26;
int n, m;
int a[N];
bool dp[N][(1 << N / 2)][N];
bool solv() {
  scanf("%d", &n);
  int qq = 0;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    if (x == 1)
      ++qq;
    else
      a[m++] = x;
  }
  n = m;
  m = qq;
  if (m < n) return false;
  if (m == 1 && n == 0) return true;
  sort(a, a + n);
  dp[0][0][m] = true;
  for (int i = 0; i < n; ++i) {
    for (int x = 0; x < (1 << i); ++x) {
      for (int q = 0; q <= m; ++q) {
        if (!dp[i][x][q]) continue;
        for (int y = x;; y = ((y - 1) & x)) {
          int s = 0;
          int z = 0;
          for (int j = 0; j < n; ++j) {
            if ((y & (1 << j))) {
              ++z;
              s += a[j];
            }
          }
          if (s > (a[i] - 1)) {
            if (y == 0) break;
            continue;
          }
          if (s + q < (a[i] - 1)) {
            if (y == 0) break;
            continue;
          }
          z += ((a[i] - 1) - s);
          if (z >= 2)
            dp[i + 1][((x ^ y) | (1 << i))][q - ((a[i] - 1) - s)] = true;
          if (y == 0) break;
        }
      }
    }
  }
  return dp[n][(1 << (n - 1))][0];
}
int main() {
  if (solv())
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

```