• if you remove a1a1, the array will look like [3,5,2][3,5,2] and it is good;
• if you remove a4a4, the array will look like [8,3,5][8,3,5] and it is good.

You have to consider all removals independently, i. e. remove the element, check if the resulting array is good, and return the element into the array.

The first line of the input contains one integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of elements in the array aa.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1061≤ai≤106) — elements of the array aa.

In the first line print one integer kk — the number of indices jj of the array aa such that after removing the jj-th element from the array it will be good (i.e. print the number of the nice indices).

In the second line print kk distinct integers j1,j2,…,jkj1,j2,…,jk in any order — nice indices of the array aa.

If there are no such indices in the array aa, just print 00 in the first line and leave the second line empty or do not print it at all.

5
2 5 1 2 2

3
4 1 5

4
8 3 5 2

2
1 4 

5
2 1 2 4 3

0

程式碼：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 2e5 + 10;
int n;
long long sum = 0;

struct Node{
    int pos;
    int val;
}node[maxn];

map<long long, int> mp;

int main() {
    scanf("%d", &n);
    mp.clear();
    for(int i = 1; i <= n; i ++) {
        int x;
        scanf("%d", &x);
        sum += x;
        node[i].pos = i;
        node[i].val = x;
        mp[x] ++;
    }

    vector<int> v;
    for(int i = 1; i <= n; i ++) {
        int temp = node[i].pos;
        mp[node[i].val] --;
        sum -= node[i].val;

        if(sum % 2 == 0 && mp[sum / 2])
            v.push_back(temp);

        sum += node[i].val;
        mp[node[i].val] ++;
    }

    printf("%d\n", v.size());
    for(int i = 0; i < v.size(); i ++) {
        printf("%d", v[i]);
        printf("%s", i != v.size() - 1 ? " " : "\n");
    }
    return 0;
}