codeforces#371 B.Filya and Homework

Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help.

Filya is given an array of non-negative integers a1, a2, ..., an. First, he pick an integer x and then he adds x to some elements of the array (no more than once), subtract x from some other elements (also, no more than once) and do no change other elements. He wants all elements of the array to be equal.

Now he wonders if it's possible to pick such integer x and change some elements of the array using this x in order to make all elements equal.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 100 000) — the number of integers in the Filya's array. The second line containsn integers a1, a2, ..., an (0 ≤ ai ≤ 109) — elements of the array.

Output

If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line of the output. Otherwise print "YES" (without quotes).

Examples

input

51 3 3 2 1

output

YES

input

51 2 3 4 5

output

NO

Note

In the first sample Filya should select x = 1, then add it to the first and the last elements of the array and subtract from the second and the third elements.

题意：给定一个序列有n个数，然后判断这些数的一些数加上或减去x，或者一些数不做处理能不能让序列的每个数最后都相同

...当时意识模糊读错题，以为x只能从这n个数中取，导致看了好久好久心态爆炸。

只有两种情况，一是把所有的数都变成这个序列的平均数，即判断小的数加上的x和大的数减去的x是否相等

二是这个序列只有两种数，无论怎么去中间值都不能得出相等的x但是这种序列可以靠小的数+（大的数-小的数） 得出

比如 1 10就可以让1+9。所以这种情况时一定可以成功的。

代码如下：

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>using namespace std;#define N 100010#define inf 2000000000000000llint a[N],b[N];int main(){    int T,i,j,n,m;    while(scanf("%d",&n)!=EOF){       for(i=1;i<=n;i++)           {            scanf("%d",&a[i]);            b[i]=a[i];           }           sort(a+1,a+1+n);           sort(b+1,b+1+n);        long long  int mid=(a[n]-a[1])/2+a[1];          int flag;         for(i=1;i<=n;i++){            if(a[i]==mid) {a[i]=0;continue;}            if(a[i]>mid){ a[i]-=mid;continue;}            else if(a[i]<mid) a[i]=mid-a[i];         }         long long int sum=0;          flag=1;         for(i=1;i<=n;i++)         {            if(a[i]==0) continue;            if(a[i]==sum) continue;            if(sum==0&&a[i]!=0) sum+=a[i];            if(sum!=0&&a[i]!=sum) {flag=0;break;}         }         //2 1000000000         int cnt=0;         for(i=1;i<=n;i++){            if(b[i]==b[1]||b[i]==b[n])                cnt++;         }         if(cnt==n) flag=1;         if(flag) {cout<<"YES"<<endl;}         else cout<<"NO"<<endl;    }    return 0;}