You are given a sequence of positive integers a1,?a2,?…,?an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).

Input
The first line contains the integer n (2?≤?n?≤?2·105). The second line contains elements of the sequence a1,?a2,?…,?an (1?≤?ai?≤?1000). All the elements are positive integers.

Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.

If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.

Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4

就是水題，從頭到尾遍歷加和，然后一個(gè)一個(gè)遍歷就好了。這個(gè)題有很多人一開(kāi)始wa了，不知道是什么情況。在判斷這個(gè)是否為平均數(shù)的時(shí)候，我才用的方法不是除以一個(gè)數(shù)，而是讓另一個(gè)數(shù)乘以一個(gè)數(shù)這樣來(lái)判斷的。因?yàn)檫@樣可以省去小數(shù)的麻煩。可能這樣就可以了。
代碼如下：

#include<iostream> #include<cstdio> #include<cmath> #include<cstring> using namespace std;const int maxx=2e5+10; int a[maxx]; int b[maxx]; int n;int main() {while(cin>>n){int sum=0;int cnt=0;for(int i=0;i<n;i++){cin>>a[i];sum+=a[i];}for(int i=0;i<n;i++){if(a[i]*(n-1)==sum-a[i]){b[cnt++]=i+1;}}cout<<cnt<<endl;for(int i=0;i<cnt;i++) cout<<b[i]<<" ";cout<<endl;} }

努力加油a啊。(^o)/~

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