Problem:
Score : 700 points

Problem Statement

There are N cats.
We number them from 1 through N.

Each of the cats wears a hat.
Cat i says: "there are exactly a_i different colors among the N - 1 hats worn by the cats except me."

Determine whether there exists a sequence of colors of the hats that is consistent with the remarks of the cats.

Constraints

2 ≤ N ≤ 10^5

1 ≤ a_i ≤ N-1

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print Yes if there exists a sequence of colors of the hats that is consistent with the remarks of the cats; print No otherwise.

Sample Input 1

3
1 2 2

Sample Output 1

Yes

For example, if cat 1, 2 and 3 wears red, blue and blue hats, respectively, it is consistent with the remarks of the cats.

Sample Input 2

3
1 1 2

Sample Output 2

No

From the remark of cat 1, we can see that cat 2 and 3 wear hats of the same color.
Also, from the remark of cat 2, we can see that cat 1 and 3 wear hats of the same color.
Therefore, cat 1 and 2 wear hats of the same color, which contradicts the remark of cat 3.

Sample Input 3

5
4 3 4 3 4

Sample Output 3

No

Sample Input 4

3
2 2 2

Sample Output 4

Yes

Sample Input 5

4
2 2 2 2

Sample Output 5

Yes

Sample Input 6

5
3 3 3 3 3

Sample Output 6

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define int long long

#define inf 1000000007LL
#define mod 1000000007LL

#define rep(i, n) for(int i = 0; i < (n); i++)
#define trep(i, n) for(int i = 0; i <= (n); i++)
#define rrep(i, n) for(int i = (n) - 1; i >= 0; i--)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define mfor(i, s, t) for(int i = (s); i < (t); i++)
#define tfor(i, s, t) for(int i = (s); i <= (t); i++)
#define rfor(i, s, t) for(int i = (t) - 1; i >= (s); i--)

int d[114514];

signed main() {
  int n;
  cin >> n;
  rep(i, n) {
    cin >> d[i];
  }
  int ma = 0;
  rep(i, n) {
    ma = max(ma, d[i]);
  }
  int f = 0;
  int h = 0;
  rep(i, n) {
    if(d[i] == ma) {
      f++;
    }
    else if(d[i] == ma - 1) {
      h++;
    }
    else {
      cout << "No" << endl;
      return 0;
    }
  }
  if(h == 0) {
    if(1 <= ma && ma <= f / 2) {
      cout << "Yes" << endl;
    }
    else if(ma == f - 1) {
      cout << "Yes" << endl;
    }
    else {
      cout << "No" << endl;
    }
  }
  else {
    if(f == 1) {
      cout << "No" << endl;
    }
    else if(h + 1 <= ma && ma <= h + f / 2) {
      cout << "Yes" << endl;
    }
    else {
      cout << "No" << endl;
    }
  }
}