#include <bits/stdc++.h>
using namespace std;
const int maxn = 25, MASK = (1 << 12) + 10;
int a[maxn], CNT;
bool dp[MASK][maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int N, n = 0, mx = 0;
  cin >> N;
  for (int i = 0; i < N; i++) {
    int x;
    cin >> x;
    if (x == 1)
      CNT++;
    else
      a[n++] = x;
    if (x == 2) return cout << "NO\n", 0;
    mx = max(mx, x);
  }
  if (CNT <= n || mx != N) return cout << "NO\n", 0;
  for (int i = 0; i <= CNT; i++) dp[0][i] = 1;
  for (int msk = 1; msk < (1 << n); msk++) {
    int id = -1, id2 = -1, C = __builtin_popcount(msk);
    for (int i = 0; i < n; i++) {
      if ((((msk) >> (i)) & 1) && (id == -1 || a[id] < a[i])) id = i;
    }
    for (int i = 0; i < n; i++) {
      if ((((msk) >> (i)) & 1) && (id2 == -1 || a[id2] < a[i]) && id != i)
        id2 = i;
    }
    for (int i = 1; i <= CNT; i++) {
      dp[msk][i] = (a[id] == C + i && (id2 == -1 || a[id] != a[id2] + 1) &&
                    dp[msk ^ (1 << id)][i]) ||
                   (dp[msk][i - 1]);
      for (int j = 1; j < i; j++) {
        int msk2 = msk;
        while (true) {
          msk2 = (msk2 - 1) & msk;
          if (msk2 == 0) break;
          dp[msk][i] |= dp[msk2][j] && dp[msk ^ msk2][i - j];
        }
      }
    }
  }
  return cout << (dp[(1 << n) - 1][CNT] ? "YES" : "NO") << endl, 0;
}

### Prompt

Generate 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 maxn = 25, MASK = (1 << 12) + 10;
int a[maxn], CNT;
bool dp[MASK][maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int N, n = 0, mx = 0;
  cin >> N;
  for (int i = 0; i < N; i++) {
    int x;
    cin >> x;
    if (x == 1)
      CNT++;
    else
      a[n++] = x;
    if (x == 2) return cout << "NO\n", 0;
    mx = max(mx, x);
  }
  if (CNT <= n || mx != N) return cout << "NO\n", 0;
  for (int i = 0; i <= CNT; i++) dp[0][i] = 1;
  for (int msk = 1; msk < (1 << n); msk++) {
    int id = -1, id2 = -1, C = __builtin_popcount(msk);
    for (int i = 0; i < n; i++) {
      if ((((msk) >> (i)) & 1) && (id == -1 || a[id] < a[i])) id = i;
    }
    for (int i = 0; i < n; i++) {
      if ((((msk) >> (i)) & 1) && (id2 == -1 || a[id2] < a[i]) && id != i)
        id2 = i;
    }
    for (int i = 1; i <= CNT; i++) {
      dp[msk][i] = (a[id] == C + i && (id2 == -1 || a[id] != a[id2] + 1) &&
                    dp[msk ^ (1 << id)][i]) ||
                   (dp[msk][i - 1]);
      for (int j = 1; j < i; j++) {
        int msk2 = msk;
        while (true) {
          msk2 = (msk2 - 1) & msk;
          if (msk2 == 0) break;
          dp[msk][i] |= dp[msk2][j] && dp[msk ^ msk2][i - j];
        }
      }
    }
  }
  return cout << (dp[(1 << n) - 1][CNT] ? "YES" : "NO") << endl, 0;
}

```