#include <bits/stdc++.h>
using namespace std;
long long sums[30000], dp[25][3000][25], can[3000], n, i, j, k, x, y, a[300];
int main() {
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  reverse(a, a + n);
  for (i = 0; i < n; i++)
    if (a[i] == 1) {
      k = i;
      break;
    }
  if (k == 0) {
    if (n == 1)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
    return 0;
  }
  if (k > 11) {
    cout << "NO" << endl;
    return 0;
  }
  for (i = 0; i < (1 << k); i++) {
    long long cnt = 0;
    for (j = 0; j < k; j++)
      if (i & (1 << j)) sums[i] += a[j], cnt++;
    can[i] = cnt;
  }
  dp[0][1][0] = 1;
  for (i = 0; i < k; i++)
    for (j = 0; j < (1 << k); j++)
      if (j & ((1 << (i + 1)) - 1) == (1 << (i + 1)) - 1)
        for (x = 0; x <= n - k; x++) {
          long long tmp = (1 << k) - 1 - j;
          for (y = tmp;; y = ((y - 1) & tmp)) {
            if (a[i] - sums[y] - 1 >= 0 && can[y] + a[i] - sums[y] - 1 != 1) {
              if (dp[i][j][x] == 1 && x + a[i] - sums[y] - 1 <= n - k &&
                  a[i] - sums[y] - 1 >= 0) {
                dp[i + 1][j | y][x + a[i] - sums[y] - 1] = 1;
              }
            }
            if (y == 0) break;
          }
        }
  if (dp[k][(1 << k) - 1][n - k])
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
long long sums[30000], dp[25][3000][25], can[3000], n, i, j, k, x, y, a[300];
int main() {
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  reverse(a, a + n);
  for (i = 0; i < n; i++)
    if (a[i] == 1) {
      k = i;
      break;
    }
  if (k == 0) {
    if (n == 1)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
    return 0;
  }
  if (k > 11) {
    cout << "NO" << endl;
    return 0;
  }
  for (i = 0; i < (1 << k); i++) {
    long long cnt = 0;
    for (j = 0; j < k; j++)
      if (i & (1 << j)) sums[i] += a[j], cnt++;
    can[i] = cnt;
  }
  dp[0][1][0] = 1;
  for (i = 0; i < k; i++)
    for (j = 0; j < (1 << k); j++)
      if (j & ((1 << (i + 1)) - 1) == (1 << (i + 1)) - 1)
        for (x = 0; x <= n - k; x++) {
          long long tmp = (1 << k) - 1 - j;
          for (y = tmp;; y = ((y - 1) & tmp)) {
            if (a[i] - sums[y] - 1 >= 0 && can[y] + a[i] - sums[y] - 1 != 1) {
              if (dp[i][j][x] == 1 && x + a[i] - sums[y] - 1 <= n - k &&
                  a[i] - sums[y] - 1 >= 0) {
                dp[i + 1][j | y][x + a[i] - sums[y] - 1] = 1;
              }
            }
            if (y == 0) break;
          }
        }
  if (dp[k][(1 << k) - 1][n - k])
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

```