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;
void dbg(const char* fmt, ...) {}
void brute(int cur, int chCnt, int accumC, int minC, int n, vector<int>& a,
           vector<int>& cnt) {
  if (cur == n) {
    printf("YES\n");
    exit(0);
  }
  if (a[cur] == accumC) {
    if (chCnt != 1) {
      cnt[a[cur]]++;
      brute(cur + 1, 0, 1, 1, n, a, cnt);
      cnt[a[cur]]--;
    }
    return;
  }
  if (a[cur] < accumC + minC) return;
  for (int j = min((a[cur] - accumC) / minC, cnt[minC]); j >= 0; --j) {
    cnt[minC] -= j;
    brute(cur, chCnt + j, accumC + minC * j, minC + 1, n, a, cnt);
    cnt[minC] += j;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  sort(a.begin(), a.end());
  if (a.back() != n) {
    printf("NO\n");
    return 0;
  }
  vector<int> cnt(n, 0);
  brute(0, 0, 1, 1, n, a, cnt);
  printf("NO\n");
  return 0;
}
