#include <bits/stdc++.h>
using namespace std;
const int N = 25, M = 12;
int n, a[N], b[N], lz[N], tl = 0, ty;
bool f[1 << M][2][N];
int lbt(int x) { return x & -x; }
int sz(int x) { return x ? sz(x ^ lbt(x)) + 1 : 0; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      ty++;
    else
      lz[tl++] = a[i];
  }
  if (ty < n / 2 + 1) return printf("NO\n"), 0;
  for (int i = 2; i <= ty; i++) f[0][1][i] = 1;
  f[0][0][1] = 1;
  for (int i = 1; i < (1 << tl); i++) {
    int ss = sz(i);
    for (int j = 1; j <= ty; j++) {
      for (int k = 0; k < tl; k++)
        if (i >> k & 1) f[i][0][j] |= f[i ^ (1 << k)][1][j] & (ss + j == lz[k]);
      for (int k = i & (i - 1); k ^ i; k = i & (k - 1))
        for (int l = 1; l < j; l++)
          f[i][1][j] |= (f[k][0][l] | f[k][1][l]) &
                        (f[k ^ i][0][j - l] | f[k ^ i][1][j - l]);
    }
  }
  printf("%s\n", f[(1 << tl) - 1][0][ty] ? "YES" : "NO");
}

### Prompt

Your task is to create 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>
using namespace std;
const int N = 25, M = 12;
int n, a[N], b[N], lz[N], tl = 0, ty;
bool f[1 << M][2][N];
int lbt(int x) { return x & -x; }
int sz(int x) { return x ? sz(x ^ lbt(x)) + 1 : 0; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      ty++;
    else
      lz[tl++] = a[i];
  }
  if (ty < n / 2 + 1) return printf("NO\n"), 0;
  for (int i = 2; i <= ty; i++) f[0][1][i] = 1;
  f[0][0][1] = 1;
  for (int i = 1; i < (1 << tl); i++) {
    int ss = sz(i);
    for (int j = 1; j <= ty; j++) {
      for (int k = 0; k < tl; k++)
        if (i >> k & 1) f[i][0][j] |= f[i ^ (1 << k)][1][j] & (ss + j == lz[k]);
      for (int k = i & (i - 1); k ^ i; k = i & (k - 1))
        for (int l = 1; l < j; l++)
          f[i][1][j] |= (f[k][0][l] | f[k][1][l]) &
                        (f[k ^ i][0][j - l] | f[k ^ i][1][j - l]);
    }
  }
  printf("%s\n", f[(1 << tl) - 1][0][ty] ? "YES" : "NO");
}

```