#include <bits/stdc++.h>
using namespace std;
const int N = 12 + 1, M = 1 << N | 5;
bool dp1[M][N + N], dp2[M][N + N];
int n, k, mx[M], c[N + N];
int main() {
  cin >> n;
  dp1[0][1] = true;
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    if (c[i] < 2) n--, k++, i--;
  }
  if (n > k) return cout << "NO", 0;
  for (int i = 1; i <= k; i++) {
    dp1[0][i] = true;
    dp2[0][i] = (i > 1);
  }
  for (int i = 0; i < n; i++) dp1[1 << i][c[i] - 1] = (c[i] > 2);
  for (int i = 1; i < 1 << n; i++) {
    int &u = mx[i] = __builtin_ctz(i);
    if (i ^ i & -i && c[u] < c[mx[i ^ i & -i]]) u = mx[i ^ i & -i];
    for (int j = 1; j <= k; j++)
      for (int x = i; x; --x &= i)
        for (int y = 0; y <= j; y++) {
          if (~x >> u & 1 && c[u] == __builtin_popcount(i) + j)
            dp1[i][j] |= dp1[x][y] && (dp1[i ^ x ^ 1 << u][j - y] ||
                                       dp2[i ^ x ^ 1 << u][j - y]);
          dp2[i][j] |= dp1[x][y] && (dp1[i ^ x][j - y] || dp2[i ^ x][j - y]);
        }
  }
  cout << (dp1[(1 << n) - 1][k] ? "YES" : "NO");
}

### 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;
const int N = 12 + 1, M = 1 << N | 5;
bool dp1[M][N + N], dp2[M][N + N];
int n, k, mx[M], c[N + N];
int main() {
  cin >> n;
  dp1[0][1] = true;
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    if (c[i] < 2) n--, k++, i--;
  }
  if (n > k) return cout << "NO", 0;
  for (int i = 1; i <= k; i++) {
    dp1[0][i] = true;
    dp2[0][i] = (i > 1);
  }
  for (int i = 0; i < n; i++) dp1[1 << i][c[i] - 1] = (c[i] > 2);
  for (int i = 1; i < 1 << n; i++) {
    int &u = mx[i] = __builtin_ctz(i);
    if (i ^ i & -i && c[u] < c[mx[i ^ i & -i]]) u = mx[i ^ i & -i];
    for (int j = 1; j <= k; j++)
      for (int x = i; x; --x &= i)
        for (int y = 0; y <= j; y++) {
          if (~x >> u & 1 && c[u] == __builtin_popcount(i) + j)
            dp1[i][j] |= dp1[x][y] && (dp1[i ^ x ^ 1 << u][j - y] ||
                                       dp2[i ^ x ^ 1 << u][j - y]);
          dp2[i][j] |= dp1[x][y] && (dp1[i ^ x][j - y] || dp2[i ^ x][j - y]);
        }
  }
  cout << (dp1[(1 << n) - 1][k] ? "YES" : "NO");
}

```