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, p[30], cnt = 0;
bool use[30];
bool dfs(int now, int cur, int son, int need, int leaf) {
  if (leaf < 0) return 0;
  if (now == cnt) return 1;
  if (!need) {
    if (son < 2) return 0;
    return dfs(now + 1, 0, 0, p[now + 1] - 1, leaf);
  }
  int i;
  for (i = cur; i < now && use[i]; ++i)
    ;
  if (p[i] > need) {
    if (son + need < 2) return 0;
    return dfs(now + 1, 0, 0, p[now + 1] - 1, leaf - need);
  }
  use[i] = 1;
  if (dfs(now, i + 1, son + 1, need - p[i], leaf)) return true;
  use[i] = 0;
  return dfs(now, i + 1, son, need, leaf);
}
int main() {
  scanf("%d", &n);
  if (n == 1) {
    puts("YES");
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    if (x > 1) p[cnt++] = x;
  }
  int leaf = n - cnt;
  if (leaf * 2 <= n) {
    puts("NO");
    return 0;
  }
  sort(p, p + cnt);
  if (p[cnt - 1] != n) {
    puts("NO");
    return 0;
  }
  puts(dfs(0, 0, 0, p[0] - 1, leaf) ? "YES" : "NO");
  return 0;
}
