#include <bits/stdc++.h>
using namespace std;
template <class T>
void print(vector<T> v) {
  cerr << "[";
  if ((int)((v).size()) != 0) cerr << v[0];
  for (__typeof((int)((v).size())) i = (1); i < ((int)((v).size())); ++i)
    cerr << "," << v[i];
  cerr << "]\n";
}
template <class T>
string i2s(T &x) {
  ostringstream o;
  o << x;
  return o.str();
}
vector<string> split(string &s, char c = ' ') {
  vector<string> all;
  int p = 0, np;
  while (np = s.find(c, p), np >= 0) {
    if (np != p) all.push_back(s.substr(p, np - p));
    p = np + 1;
  }
  if (p < (int)((s).size())) all.push_back(s.substr(p));
  return all;
}
int n;
vector<int> d;
void readData() {
  scanf("%d", &n);
  d.resize(n);
  for (__typeof(n) i = (0); i < (n); ++i) scanf("%d", &(d[i]));
  sort(d.begin(), d.end(), greater<int>());
}
vector<int> r;
vector<int> par;
int slack;
int _solve(int ind) {
  if (d[ind] == 1) {
    return 1;
  }
  int start = 0;
  if (ind > 1 && d[ind - 1] == d[ind]) start = par[ind - 1];
  for (int p = start; p < ind; p++) {
    if (r[p] > d[ind] + 1 || (r[p] == d[ind] + 1 && r[p] != d[p])) {
      r[p] -= d[ind];
      par[ind] = p;
      if (_solve(ind + 1)) return 1;
      r[p] += d[ind];
    }
  }
  return 0;
}
int solve() {
  for (__typeof(n) i = (0); i < (n); ++i)
    if (d[i] == 2) return 0;
  if (n == 1 && d[0] == 1) return 1;
  if (d[0] != n) return 0;
  r = d;
  par = vector<int>(n);
  slack = 0;
  return _solve(1);
}
int main(int argc, char *argv[]) {
  readData();
  if (solve())
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
void print(vector<T> v) {
  cerr << "[";
  if ((int)((v).size()) != 0) cerr << v[0];
  for (__typeof((int)((v).size())) i = (1); i < ((int)((v).size())); ++i)
    cerr << "," << v[i];
  cerr << "]\n";
}
template <class T>
string i2s(T &x) {
  ostringstream o;
  o << x;
  return o.str();
}
vector<string> split(string &s, char c = ' ') {
  vector<string> all;
  int p = 0, np;
  while (np = s.find(c, p), np >= 0) {
    if (np != p) all.push_back(s.substr(p, np - p));
    p = np + 1;
  }
  if (p < (int)((s).size())) all.push_back(s.substr(p));
  return all;
}
int n;
vector<int> d;
void readData() {
  scanf("%d", &n);
  d.resize(n);
  for (__typeof(n) i = (0); i < (n); ++i) scanf("%d", &(d[i]));
  sort(d.begin(), d.end(), greater<int>());
}
vector<int> r;
vector<int> par;
int slack;
int _solve(int ind) {
  if (d[ind] == 1) {
    return 1;
  }
  int start = 0;
  if (ind > 1 && d[ind - 1] == d[ind]) start = par[ind - 1];
  for (int p = start; p < ind; p++) {
    if (r[p] > d[ind] + 1 || (r[p] == d[ind] + 1 && r[p] != d[p])) {
      r[p] -= d[ind];
      par[ind] = p;
      if (_solve(ind + 1)) return 1;
      r[p] += d[ind];
    }
  }
  return 0;
}
int solve() {
  for (__typeof(n) i = (0); i < (n); ++i)
    if (d[i] == 2) return 0;
  if (n == 1 && d[0] == 1) return 1;
  if (d[0] != n) return 0;
  r = d;
  par = vector<int>(n);
  slack = 0;
  return _solve(1);
}
int main(int argc, char *argv[]) {
  readData();
  if (solve())
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

```