#include <bits/stdc++.h>
const int inf = 100100;
const int maxn = (int)1e5 + 10;
const int mod = (int)1e9 + 7;
using namespace std;
int n, m;
int a[25];
int b[25];
int dp[13][1 << 13][24];
int get(int p, int mask, int c) {
  if (p == m) {
    return mask == (1 << (m - 1)) && c == 0;
  }
  int &res = dp[p][mask][c];
  if (res != -1) return res;
  res = 0;
  for (int h = mask;; h = ((h - 1) & mask)) {
    int d = b[p] - 1;
    for (int i = 0; i < p; i++) {
      if (h & (1 << i)) {
        if (b[i] + 1 == b[p]) d = -1;
        d -= b[i];
      }
    }
    if (c >= d && d >= 0) {
      res |= get(p + 1, (mask ^ h) | (1 << p), c - d);
    }
    if (h == 0) break;
  }
  return res;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a[x]++;
  }
  if (n == 1) {
    cout << "YES\n";
    return;
  }
  if (a[1] * 2 < n + 1) {
    cout << "NO\n";
    return;
  }
  if (a[1] < 2 || a[n] != 1 || a[2] > 0) {
    cout << "NO\n";
    return;
  }
  if (a[n - 1] > 0) {
    cout << "NO\n";
    return;
  }
  m = 0;
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j < a[i]; j++) b[m++] = i;
  }
  memset(dp, -1, sizeof dp);
  if (get(0, 0, a[1])) {
    cout << "YES\n";
  } else
    cout << "NO\n";
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
const int inf = 100100;
const int maxn = (int)1e5 + 10;
const int mod = (int)1e9 + 7;
using namespace std;
int n, m;
int a[25];
int b[25];
int dp[13][1 << 13][24];
int get(int p, int mask, int c) {
  if (p == m) {
    return mask == (1 << (m - 1)) && c == 0;
  }
  int &res = dp[p][mask][c];
  if (res != -1) return res;
  res = 0;
  for (int h = mask;; h = ((h - 1) & mask)) {
    int d = b[p] - 1;
    for (int i = 0; i < p; i++) {
      if (h & (1 << i)) {
        if (b[i] + 1 == b[p]) d = -1;
        d -= b[i];
      }
    }
    if (c >= d && d >= 0) {
      res |= get(p + 1, (mask ^ h) | (1 << p), c - d);
    }
    if (h == 0) break;
  }
  return res;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a[x]++;
  }
  if (n == 1) {
    cout << "YES\n";
    return;
  }
  if (a[1] * 2 < n + 1) {
    cout << "NO\n";
    return;
  }
  if (a[1] < 2 || a[n] != 1 || a[2] > 0) {
    cout << "NO\n";
    return;
  }
  if (a[n - 1] > 0) {
    cout << "NO\n";
    return;
  }
  m = 0;
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j < a[i]; j++) b[m++] = i;
  }
  memset(dp, -1, sizeof dp);
  if (get(0, 0, a[1])) {
    cout << "YES\n";
  } else
    cout << "NO\n";
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```