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 double eps = 1e-6;
const int N = 1e3 + 7;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
int a[N], b[N], c[N], n, cnt, sum, found = 0;
bool cmp(int a, int b) { return a > b; }
void Dfs(int x) {
  if (x == n) {
    if (sum == cnt) found = 1;
    return;
  }
  for (int i = 0; i < x; i++)
    if (a[i] > b[x] && b[i] != b[x] + 1) {
      a[i] -= b[x];
      sum -= b[x];
      c[i]++;
      Dfs(x + 1);
      sum += b[x];
      c[i]--;
      a[i] += b[x];
    }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (n == 1 && a[0] == 1) {
    puts("YES");
    return 0;
  }
  sort(a, a + n, cmp);
  for (int i = 0; i < n; i++) b[i] = a[i];
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] != 1) break;
    cnt++;
  }
  for (int i = 0; i < n; i++)
    if (a[i] == 2) {
      puts("NO");
      return 0;
    }
  if (cnt * 2 <= n) {
    puts("NO");
    return 0;
  }
  n -= cnt;
  for (int i = 0; i < n; i++) sum += a[i] - 1;
  Dfs(1);
  if (found)
    puts("YES");
  else
    puts("NO");
}
