#include <bits/stdc++.h>
using namespace std;
inline void setmin(int &x, int y) {
  if (y < x) x = y;
}
inline void setmax(int &x, int y) {
  if (y > x) x = y;
}
inline void setmin(long long &x, long long y) {
  if (y < x) x = y;
}
inline void setmax(long long &x, long long y) {
  if (y > x) x = y;
}
const int N = 100000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
int sum[1 << 11];
bool dp[11 + 1][1 << 11];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  int n;
  cin >> n;
  vector<int> a;
  int cnt1 = 0;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x != 1)
      a.push_back(x);
    else
      cnt1++;
  }
  sort(a.begin(), a.end());
  if (cnt1 * 2 <= n || (int)a.size() > 0 && a.back() != n) {
    cout << "NO\n";
    return 0;
  }
  if ((int)a.size() == 0) {
    if (n == 1)
      cout << "YES\n";
    else
      cout << "NO\n";
    return 0;
  }
  n = (int)a.size();
  for (int mask = 0; mask < (1 << n); mask++) {
    sum[mask] = 0;
    for (int i = 0; i < n; i++)
      if ((mask >> i) & 1) sum[mask] += a[i];
  }
  dp[0][0] = true;
  for (int i = 0; i < n; i++)
    for (int mask = 0; mask < (1 << n); mask++) {
      if (!dp[i][mask]) continue;
      for (int sm = mask;; sm = (sm - 1) & mask) {
        if (sum[sm] + 1 <= a[i] &&
            __builtin_popcount(sm) + (a[i] - 1 - sum[sm]) >= 2)
          dp[i + 1][mask ^ sm | (1 << i)] = true;
        if (sm == 0) break;
      }
    }
  cout << (dp[n][1 << (n - 1)] ? "YES\n" : "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;
inline void setmin(int &x, int y) {
  if (y < x) x = y;
}
inline void setmax(int &x, int y) {
  if (y > x) x = y;
}
inline void setmin(long long &x, long long y) {
  if (y < x) x = y;
}
inline void setmax(long long &x, long long y) {
  if (y > x) x = y;
}
const int N = 100000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
int sum[1 << 11];
bool dp[11 + 1][1 << 11];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  int n;
  cin >> n;
  vector<int> a;
  int cnt1 = 0;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x != 1)
      a.push_back(x);
    else
      cnt1++;
  }
  sort(a.begin(), a.end());
  if (cnt1 * 2 <= n || (int)a.size() > 0 && a.back() != n) {
    cout << "NO\n";
    return 0;
  }
  if ((int)a.size() == 0) {
    if (n == 1)
      cout << "YES\n";
    else
      cout << "NO\n";
    return 0;
  }
  n = (int)a.size();
  for (int mask = 0; mask < (1 << n); mask++) {
    sum[mask] = 0;
    for (int i = 0; i < n; i++)
      if ((mask >> i) & 1) sum[mask] += a[i];
  }
  dp[0][0] = true;
  for (int i = 0; i < n; i++)
    for (int mask = 0; mask < (1 << n); mask++) {
      if (!dp[i][mask]) continue;
      for (int sm = mask;; sm = (sm - 1) & mask) {
        if (sum[sm] + 1 <= a[i] &&
            __builtin_popcount(sm) + (a[i] - 1 - sum[sm]) >= 2)
          dp[i + 1][mask ^ sm | (1 << i)] = true;
        if (sm == 0) break;
      }
    }
  cout << (dp[n][1 << (n - 1)] ? "YES\n" : "NO\n");
  return 0;
}

```