Codeforces #371（Div.2）B. Filya and Homework【思维】

B. Filya and Homework

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., a_n. First, he pick an integerx 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 thisx 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, ..., a_n (0 ≤ a_i ≤ 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.

题目大意：

给你一个长度为n的序列，问你能否找到一个值x，使得序列中的任意数值可以-x或者+x的情况下，使得序列中的数最终变成相同的数。

思路：

分类讨论：

①如果序列中的数字的种类大于了3，那么显然不可能得到期望序列，即输出NO

②如果序列中的数字的种类小于等于2，那么显然一定可以得到期望序列，即输出YES

③如果序列中的数字的种类邓雨了3，那么最小的数字加上最大的数字等于中间值的二倍，明显也是YES的情况，否则为NO。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<map>using namespace std;int a[1000050];int b[1000050];int main(){    int n;    while(~scanf("%d",&n))    {        map<int ,int >s;        int cont=1;        int contz=0;        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);            if(s[a[i]]==0)            {                s[a[i]]=cont++;                b[contz++]=a[i];            }        }        sort(b,b+contz);        if(cont>4)        {            printf("NO\n");fflush(stdout);        }        else        {            if(cont==4)            {                if(b[0]+b[2]==b[1]*2)                {                    printf("YES\n");fflush(stdout);                }                else                {                    printf("NO\n");fflush(stdout);                }            }            if(cont==3||cont==2)            {                printf("YES\n");fflush(stdout);            }        }    }}