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++ (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define good {puts("No");return 0;}
#define Hao {puts("Yes");return 0;}
using namespace std;

const int S=(1<<20)+5;
char buf[S],*H,*T;
inline char Get()
{
    if(H==T) T=(H=buf)+fread(buf,1,S,stdin);
    if(H==T) return -1;return *H++;
}
inline int read()
{
    int x=0;char c=Get();
    while(!isdigit(c)) c=Get();
    while(isdigit(c)) x=x*10+c-'0',c=Get();
    return x;
}

inline void upmin(int &x,const int &y){if(y<x) x=y;}
inline void upmax(int &x,const int &y){if(y>x) x=y;}

int a[100010],n;

int main()
{
    n=read();
    int mn=n+1,mx=0;
    for(int i=1;i<=n;i++)
    {
        a[i]=read();
        upmin(mn,a[i]);
        upmax(mx,a[i]);
    }
    if(mx>mn+1) good
    if(mx==mn)
    {
        if(mn==n-1) Hao
        if(2*mn<=n) Hao
        else good
    }
    else
    {
        int cnt=0;
        for(int i=1;i<=n;i++) cnt+=(a[i]==mn);
        if(mn<cnt) good
        if(n-cnt<2*(mx-cnt)) good
        else Hao
    }
    return 0;
}