#include <bits/stdc++.h>
using namespace std;
int c[25];
bool dp[2050][25][3] = {false};
int cnt1bit(int s) {
  int c = 0;
  while (s) c += s & 1, s >>= 1;
  return c;
}
int main() {
  int n;
  cin >> n;
  int leafCnt = 0;
  int it = 0;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    if (a == 1)
      leafCnt++;
    else
      c[it++] = a;
  }
  int restNode = n - leafCnt;
  if (restNode > 11) {
    puts("NO");
    return 0;
  }
  dp[0][1][0] = true;
  for (int i = 2; i <= leafCnt; i++) dp[0][i][1] = true;
  for (int S = 1; S < (1 << restNode); S++) {
    for (int k = 1; k <= leafCnt; k++)
      if (dp[S][k - 1][0] || dp[S][k - 1][1]) dp[S][k][1] = true;
    int ps = cnt1bit(S);
    for (int x = 0; x < restNode; x++) {
      if (!(S & (1 << x))) continue;
      int L = c[x] - ps;
      if (L >= 0 && L <= leafCnt && dp[S ^ (1 << x)][L][1]) dp[S][L][0] = true;
    }
    for (int S1 = 0; S1 <= S; S1++) {
      if (S1 & (~S)) continue;
      for (int k = 0; k <= leafCnt; k++) {
        for (int k1 = 0; k1 <= k; k1++) {
          if (dp[S1][k1][0] && (dp[S ^ S1][k - k1][0] || dp[S ^ S1][k - k1][1]))
            dp[S][k][1] = true;
        }
      }
    }
  }
  if (dp[(1 << restNode) - 1][leafCnt][0])
    puts("YES");
  else
    puts("NO");
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 c[25];
bool dp[2050][25][3] = {false};
int cnt1bit(int s) {
  int c = 0;
  while (s) c += s & 1, s >>= 1;
  return c;
}
int main() {
  int n;
  cin >> n;
  int leafCnt = 0;
  int it = 0;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    if (a == 1)
      leafCnt++;
    else
      c[it++] = a;
  }
  int restNode = n - leafCnt;
  if (restNode > 11) {
    puts("NO");
    return 0;
  }
  dp[0][1][0] = true;
  for (int i = 2; i <= leafCnt; i++) dp[0][i][1] = true;
  for (int S = 1; S < (1 << restNode); S++) {
    for (int k = 1; k <= leafCnt; k++)
      if (dp[S][k - 1][0] || dp[S][k - 1][1]) dp[S][k][1] = true;
    int ps = cnt1bit(S);
    for (int x = 0; x < restNode; x++) {
      if (!(S & (1 << x))) continue;
      int L = c[x] - ps;
      if (L >= 0 && L <= leafCnt && dp[S ^ (1 << x)][L][1]) dp[S][L][0] = true;
    }
    for (int S1 = 0; S1 <= S; S1++) {
      if (S1 & (~S)) continue;
      for (int k = 0; k <= leafCnt; k++) {
        for (int k1 = 0; k1 <= k; k1++) {
          if (dp[S1][k1][0] && (dp[S ^ S1][k - k1][0] || dp[S ^ S1][k - k1][1]))
            dp[S][k][1] = true;
        }
      }
    }
  }
  if (dp[(1 << restNode) - 1][leafCnt][0])
    puts("YES");
  else
    puts("NO");
  return 0;
}

```