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 = 24;
int N, A[MAXN], use[MAXN], c;
bool dfs(int p, int s, int k) {
  if (p == N) return c == N - 1;
  if (k == 0) return dfs(p + 1, p, A[p + 1] - 1);
  for (int i = s; i >= 0; i--) {
    if (use[i] || A[i] > k || A[i] == A[p] - 1) continue;
    use[i] = 1;
    c++;
    if (dfs(p, i - 1, k - A[i])) return 1;
    use[i] = 0;
    c--;
  }
  return 0;
}
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> A[i];
  sort(A, A + N);
  bool res = dfs(0, -1, A[0] - 1);
  cout << (res ? "YES\n" : "NO\n");
}
