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;
bool isok = false;
set<vector<int> > se;
set<vector<int> > s2;
vector<vector<int> > nx;
void get(vector<int> v, vector<int> n, int mok) {
  n.push_back(mok);
  if (s2.find(n) != s2.end()) return;
  s2.insert(n);
  n.pop_back();
  if (mok < 0) return;
  if (v.empty()) {
    if (mok == 0) {
      nx.push_back(n);
    }
    return;
  }
  int x = v[v.size() - 1];
  v.pop_back();
  get(v, n, mok - x);
  n.push_back(x);
  get(v, n, mok);
}
void calc(vector<int> now, vector<int> nd) {
  if (nd.empty()) {
    if (now.size() == 1) isok = true;
    return;
  }
  if (se.find(now) != se.end()) return;
  se.insert(now);
  int x = nd[nd.size() - 1];
  nd.pop_back();
  vector<vector<int> > nxt;
  nx.clear();
  vector<int> ddd;
  s2.clear();
  get(now, ddd, x - 1);
  nxt = nx;
  for (int i = 0; i < nxt.size(); i++) {
    if (nxt[i].size() == now.size() - 1) continue;
    nxt[i].push_back(x);
    sort(nxt[i].begin(), nxt[i].end());
    calc(nxt[i], nd);
  }
}
int main() {
  int num;
  scanf("%d", &num);
  vector<int> v1, v2;
  for (int i = 0; i < num; i++) {
    int z;
    scanf("%d", &z);
    if (z == 1)
      v1.push_back(z);
    else
      v2.push_back(z);
  }
  sort(v2.begin(), v2.end());
  reverse(v2.begin(), v2.end());
  calc(v1, v2);
  if (isok)
    printf("YES\n");
  else
    printf("NO\n");
}
