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 = 3e1 + 4;
int n, num[maxn];
bool vis[maxn];
int dp[maxn];
int check(int pos, int res) {
  memset(dp, 0, sizeof dp);
  for (int i = pos; i >= 1; --i) {
    if (vis[i]) continue;
    for (int j = res; j >= num[i]; --j)
      dp[j] = max(dp[j], dp[j - num[i]] + num[i]);
    if (dp[res] == res) return i;
  }
  return -1;
}
int main() {
  ios::sync_with_stdio(false);
  int i, j, k, kase;
  kase = 1;
  while (kase--) {
    cin >> n;
    for (i = 1; i <= n; ++i) cin >> num[i];
    sort(num + 1, num + 1 + n);
    memset(vis, false, sizeof vis);
    bool flag = true;
    for (i = 1; i <= n && flag; ++i) {
      int res = num[i] - 1;
      for (j = i - 1; j > 0; --j)
        if (num[j] < num[i] - 1) break;
      while (res != 0) {
        int ans = check(j, res);
        if (ans == -1) {
          flag = false;
          break;
        }
        vis[ans] = true;
        res -= num[ans];
      }
    }
    for (i = 1; i < n; ++i)
      if (!vis[i]) {
        flag = false;
        break;
      }
    if (flag)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
