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>
#define ll int
#define pb push_back
#define mp make_pair
#define ld long double
#define F first
#define S second
#define pii pair<ll,ll>

using namespace :: std;

const ll maxn=1e6+500;
const ll mod=998244353;
const ll inf=1e9+500;

ll c[maxn];

void good(){
    cout<<"Yes";
    exit(0);
}
void bad(){
    cout<<"No";
    exit(0);
}
int main()
{
    ll n;
    cin>>n;
    ll mi=n;
    ll ma=0;
    for(ll i=0;i<n;i++){
        ll v;
        cin>>v;
        c[v]++;
        mi=min(mi,v);
        ma=max(ma,v);
    }
    if(ma-mi>1){
        bad();
    }
    if(ma==mi){
        if(ma==n-1){
            good();
        }
        else{
            if(ma*2<=n){
                good();
            }else{
                bad();
            }
        }
    }
    if(c[mi]>=ma){
        bad();
    }
    if(c[mi]+(ma-c[mi])*2>n){
        bad();
    }
    good();
}