#include <bits/stdc++.h>
using namespace std;
int a[30];
int dp[580][25];
bool flag;
int vis[30];
int n;
void dfs(int pos) {
  int b[30];
  if (flag) return;
  int i, j;
  if (pos == n) {
    for (i = 0; i < n - 1; i++) {
      if (vis[i] == 0) return;
    }
    flag = true;
    return;
  }
  if (a[pos] == 1) {
    dfs(pos + 1);
  }
  int num = 0;
  int ss = 0;
  for (i = 0; i < pos; i++) {
    if (vis[i] == 0) b[num++] = i, ss += a[i];
  }
  int M = (1 << num);
  for (i = 0; i < M; i++) {
    int sum = 0;
    int g = 0;
    for (j = 0; j < num; j++) {
      if ((1 << j) & i) sum += a[b[j]], g++;
    }
    if (g <= 1) continue;
    if (sum == a[pos] - 1) {
      for (j = 0; j < num; j++) {
        if ((1 << j) & i) {
          vis[b[j]] = 1;
        }
      }
      dfs(pos + 1);
      for (j = 0; j < num; j++) {
        if ((1 << j) & i) vis[b[j]] = 0;
      }
    }
  }
  return;
}
int main() {
  int i, j;
  while (cin >> n) {
    bool judge = false;
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      if (a[i] != 1) judge = true;
    }
    if (judge == false && n > 1) {
      puts("NO");
      continue;
    }
    sort(a, a + n);
    for (i = 0; i < n; i++) vis[i] = 0;
    flag = false;
    dfs(0);
    if (flag == false)
      puts("NO");
    else
      puts("YES");
  }
  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 a[30];
int dp[580][25];
bool flag;
int vis[30];
int n;
void dfs(int pos) {
  int b[30];
  if (flag) return;
  int i, j;
  if (pos == n) {
    for (i = 0; i < n - 1; i++) {
      if (vis[i] == 0) return;
    }
    flag = true;
    return;
  }
  if (a[pos] == 1) {
    dfs(pos + 1);
  }
  int num = 0;
  int ss = 0;
  for (i = 0; i < pos; i++) {
    if (vis[i] == 0) b[num++] = i, ss += a[i];
  }
  int M = (1 << num);
  for (i = 0; i < M; i++) {
    int sum = 0;
    int g = 0;
    for (j = 0; j < num; j++) {
      if ((1 << j) & i) sum += a[b[j]], g++;
    }
    if (g <= 1) continue;
    if (sum == a[pos] - 1) {
      for (j = 0; j < num; j++) {
        if ((1 << j) & i) {
          vis[b[j]] = 1;
        }
      }
      dfs(pos + 1);
      for (j = 0; j < num; j++) {
        if ((1 << j) & i) vis[b[j]] = 0;
      }
    }
  }
  return;
}
int main() {
  int i, j;
  while (cin >> n) {
    bool judge = false;
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      if (a[i] != 1) judge = true;
    }
    if (judge == false && n > 1) {
      puts("NO");
      continue;
    }
    sort(a, a + n);
    for (i = 0; i < n; i++) vis[i] = 0;
    flag = false;
    dfs(0);
    if (flag == false)
      puts("NO");
    else
      puts("YES");
  }
  return 0;
}

```