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 nmax = 25;
int n, arr[nmax];
int used[nmax];
void rec(int node) {
  if (node == 0) {
    for (int i = 1; i <= n; i++)
      if (arr[i]) return;
    cout << "YES" << endl;
    exit(0);
  }
  for (int p = n; p > node; p--)
    if ((arr[p] >= arr[node] + 2) || (used[p] && arr[p] >= arr[node] + 1)) {
      int mem_p = arr[p];
      arr[p] = arr[p] - arr[node] - 1;
      used[p]++;
      rec(node - 1);
      arr[p] = mem_p;
      used[p]--;
    }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  sort(arr + 1, arr + n + 1);
  for (int i = 1; i <= n; i++)
    if (arr[i] == 2) {
      cout << "NO" << endl;
      return 0;
    }
  for (int i = 1; i <= n; i++) arr[i]--;
  rec(n - 1);
  cout << "NO" << endl;
  return 0;
}
