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)

//#define _CRT_SECURE_NO_WARNINGS
//#include <bits/stdc++.h>
#include "bits/stdc++.h"

#define rep(i,n) for(ll (i)=0;(i)<(ll)(n);(i)++)
#define all(x) (x).begin(),(x).end()

#define MOD 1000000007LL
#define INF (1LL<<60LL)
#define int long long
#define MAX_N (100001)
typedef long long ll;

using namespace std;


signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n; cin >> n;
	vector<int> v(n);
	rep(i, n) cin >> v[i];
	map<int, int> m_1;
	map<int, int> m_2;
	int ans = 0;
	rep(i, n) {
		int k1 = -v[i] + i;
		int k2 = v[i] + i;
		if (m_1.find(k2) != m_1.end())ans += m_1[k2];
		if (m_1.find(k1) == m_1.end()) m_1[k1] = 1;
		else m_1[k1]++;
	}
	rep(i, n) {
		int k1 = -v[i] - i;
		int k2 = v[i] - i;
		//cout << k1  << " "<< k2 << endl;
		if (m_2.find(k2) != m_2.end())ans += m_2[k2];
		if (m_2.find(k1) == m_2.end()) m_2[k1] = 1;
		else m_2[k1]++;
	}
	cout << ans << endl;
}