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 a[30];
void solve(vector<pair<int, int> > &v, int k);
void dfs(vector<pair<int, int> > &v, int sum, int k, int cnt = 0,
         int step = 0) {
  if (sum < 0) return;
  if (step == (int)v.size()) {
    if (sum) return;
    if (cnt < 2) return;
    if (v.back().first == a[k - 1])
      v.back().second++;
    else
      v.push_back(make_pair(a[k - 1], 1));
    solve(v, k - 1);
    v.back().second--;
    if (!v.back().second) v.pop_back();
    return;
  }
  for (int i = 0; i <= v[step].second; ++i) {
    v[step].second -= i;
    dfs(v, sum - i * v[step].first, k, cnt + i, step + 1);
    v[step].second += i;
  }
}
void solve(vector<pair<int, int> > &v, int k) {
  if (!k) {
    int cnt = 0;
    for (int i = 0; i < (int)v.size(); ++i) cnt += (v[i].second != 0);
    if (cnt != 1) return;
    cout << "YES\n";
    exit(0);
  }
  dfs(v, a[k - 1] - 1, k);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  sort(a, a + n);
  reverse(a, a + n);
  int k = n;
  vector<pair<int, int> > v(0);
  while (k && a[k - 1] == 1) --k;
  v.push_back(make_pair(1, n - k));
  solve(v, k);
  cout << "NO\n";
}
