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 <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) c.begin(), c.end()
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

int main() {
    int n;
    cin >> n;
    VI a(n);
    REP(i,n) scanf("%d", &a[i]);
    sort(ALL(a));

    bool pos = false;

    if (a[0] == a[n-1]){
        if (a[0] == n-1 || 2*a[0] <= n) pos = true;
    }else if (a[0] == a[n-1] - 1){
        int x = 0;
        REP(i,n) x += (a[i] == a[0]);
        int y = a[n-1] - x;
        if (y > 0 && n - x >= 2 * y) pos = true;
    }

    cout << (pos ? "Yes" : "No") << endl;

    return 0;
}