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
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double pi = 3.14159265354;
int t, n, m, k, cnt;
int a[maxn];
bool vis[maxn];
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  cin >> n;
  int sum = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    sum += a[i];
  }
  sort(a + 1, a + 1 + n);
  if (sum == n * 3) {
    cout << "YES" << endl;
    return 0;
  }
  vis[n] = 1;
  for (int i = 1; i <= n; ++i) {
    int cnt = 1;
    for (int j = i - 1; j >= 1 && a[i] > cnt; --j) {
      if (!vis[j]) {
        if (a[j] + 2 <= a[i] && cnt + a[j] <= a[i]) {
          cnt += a[j];
          vis[j] = 1;
        }
      }
    }
    if (a[i] != cnt) {
      cout << "NO" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (!vis[i]) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  return 0;
}
