Problem:
Score: 500 points

Problem Statement

You are the top spy of AtCoder Kingdom. To prevent the stolen secret from being handed to AlDebaran Kingdom, you have sneaked into the party where the transaction happens.

There are N attendees in the party, and they are given attendee numbers from 1 through N. The height of Attendee i is A_i.

According to an examination beforehand, you know that a pair of attendees satisfying the condition below will make the transaction.

The absolute difference of their attendee numbers is equal to the sum of their heights.

There are \frac{N(N-1)}{2} ways to choose two from the N attendees and make a pair. Among them, how many satisfy the condition above?

P.S.: We cannot let you know the secret.

Constraints

All values in input are integers.

2 \leq N \leq 2 \times 10^5

1 \leq A_i \leq 10^9\ (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \dots A_N

Output

Print the number of pairs satisfying the condition.

Sample Input 1

6
2 3 3 1 3 1

Sample Output 1

3

A_1 + A_4 = 3, so the pair of Attendee 1 and 4 satisfy the condition.

A_2 + A_6 = 4, so the pair of Attendee 2 and 6 satisfy the condition.

A_4 + A_6 = 2, so the pair of Attendee 4 and 6 satisfy the condition.

No other pair satisfies the condition, so you should print 3.

Sample Input 2

6
5 2 4 2 8 8

Sample Output 2

0

No pair satisfies the condition, so you should print 0.

Sample Input 3

32
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5

Sample Output 3

22

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//  Copyright © 2020 Diego Garcia Rodriguez del Campo. All rights reserved.
#include<bits/stdc++.h>
#define MAX 200005
#define optimiza_io cin.tie(0); ios_base::sync_with_stdio(0);
#define what_is(x) cerr << #x << " is " << x << endl;
using namespace std;
typedef long long ll;

ll N, a[MAX], can[MAX], ans;

int main()
{
    optimiza_io
    cin >> N;
    for( ll i = 1; i <= N; i ++ ) {
        cin >> a[i];
        ll p = i - a[i] + 1;
        if( p >= 1 )
            can[p]++;
    }
    for( ll i = 1; i <= N; i ++ ) {
        ll p = i + a[i] + 1;
        if( p <= N )
            ans += can[p];
    }
    
    cout << ans;
    return 0;
}

// CHECAR LIMITES