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;
int n;
int a[233], vis[223], fa[233], son[233], b[233];
bool check() {
  for (int i = 1; i < n; i++)
    if (son[i] < 2 && b[i] != 1) return false;
  return true;
}
int T = 1e7;
bool dfs(int x) {
  if (clock() / double(CLOCKS_PER_SEC) > .8) {
    puts("NO");
    exit(0);
  }
  if (x == n) {
    return check();
  }
  for (int i = x + 1; i <= n; i++)
    if (b[i] - a[x] >= 1) {
      if (b[i] - a[x] == 1 && son[i] == 0) continue;
      b[i] -= a[x];
      son[i]++;
      if (dfs(x + 1)) return true;
      son[i]--;
      b[i] += a[x];
    }
  return false;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = a[i];
    if (a[i] == 2) return puts("NO"), 0;
  }
  sort(a + 1, a + 1 + n);
  sort(b + 1, b + 1 + n);
  puts(dfs(1) ? "YES" : "NO");
  return 0;
}
