#include <bits/stdc++.h>
using namespace std;
using LL = int64_t;
const int INF = 0x3f3f3f3f;
const LL LINF = 0x3f3f3f3f3f3f3f3f;
const double EPS = 1e-9;
const double PI = 2 * asin(1);
const int MX = 24 + 2;
int g_n;
int vi_n[MX];
vector<bool> vb;
void pretreat() {}
bool input() {
  cin >> g_n;
  if (cin.eof()) return false;
  vb.clear(), vb.resize(g_n + 1);
  for (int i = 1; i <= g_n; ++i) scanf(" %d", vi_n + i);
  return true;
}
bool dfs(int u, int sum, int son) {
  if (u == g_n + 1) return true;
  if (sum + 1 == vi_n[u]) {
    if (son == 1)
      return false;
    else
      return dfs(u + 1, 0, 0);
  }
  int t = -1;
  for (int i = u + 1; i <= g_n; ++i) {
    if (vb[i] || vi_n[i] + sum > vi_n[u] || vi_n[i] == t) continue;
    vb[i] = true;
    t = vi_n[i];
    if (dfs(u, sum + vi_n[i], son + 1)) return true;
    vb[i] = false;
  }
  return false;
}
void solve() {
  sort(vi_n + 1, vi_n + 1 + g_n);
  reverse(vi_n + 1, vi_n + 1 + g_n);
  bool f = true, f1 = false;
  for (int i = 1; i <= g_n && f; ++i) {
    if (vi_n[i] == 2) f = false;
    if (vi_n[i] == 1) f1 = true;
  }
  if (!(f1 && f)) {
    puts("NO");
    return;
  }
  if (vi_n[1] == g_n && dfs(1, 0, 0))
    puts("YES");
  else
    puts("NO");
}
int main() {
  pretreat();
  while (input()) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to 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;
using LL = int64_t;
const int INF = 0x3f3f3f3f;
const LL LINF = 0x3f3f3f3f3f3f3f3f;
const double EPS = 1e-9;
const double PI = 2 * asin(1);
const int MX = 24 + 2;
int g_n;
int vi_n[MX];
vector<bool> vb;
void pretreat() {}
bool input() {
  cin >> g_n;
  if (cin.eof()) return false;
  vb.clear(), vb.resize(g_n + 1);
  for (int i = 1; i <= g_n; ++i) scanf(" %d", vi_n + i);
  return true;
}
bool dfs(int u, int sum, int son) {
  if (u == g_n + 1) return true;
  if (sum + 1 == vi_n[u]) {
    if (son == 1)
      return false;
    else
      return dfs(u + 1, 0, 0);
  }
  int t = -1;
  for (int i = u + 1; i <= g_n; ++i) {
    if (vb[i] || vi_n[i] + sum > vi_n[u] || vi_n[i] == t) continue;
    vb[i] = true;
    t = vi_n[i];
    if (dfs(u, sum + vi_n[i], son + 1)) return true;
    vb[i] = false;
  }
  return false;
}
void solve() {
  sort(vi_n + 1, vi_n + 1 + g_n);
  reverse(vi_n + 1, vi_n + 1 + g_n);
  bool f = true, f1 = false;
  for (int i = 1; i <= g_n && f; ++i) {
    if (vi_n[i] == 2) f = false;
    if (vi_n[i] == 1) f1 = true;
  }
  if (!(f1 && f)) {
    puts("NO");
    return;
  }
  if (vi_n[1] == g_n && dfs(1, 0, 0))
    puts("YES");
  else
    puts("NO");
}
int main() {
  pretreat();
  while (input()) {
    solve();
  }
  return 0;
}

```