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)

/*
author: madhav_1999 aka orthodoxparadox
3 May 2020 at 5:44 PM
*/

#include <bits/stdc++.h>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define dbl long double
#define int ll
#define ll long long
#define pii pair<int, int>
#define len(x) (int) x.size()
#define rev(a) reverse(all(a))
#define oiint ostream_iterator<int>(cout, " ") 
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define initialise(a, x) memset(a, x, sizeof(a))
#define onlyunique(v) v.erase(unique(all(v)), v.end()); //only for sorted vector
#define inf 2e18
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _CRT_SECURE_NO_WARNINGS
#ifdef MADHAV
#define dbg(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){cerr << name << " : " << arg1 << endl;}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');
	cerr.write(names, comma - names) << " : " << arg1 << "  ";
	__f(comma + 1, args...);}
#else
#define dbg(...) 42
#endif
template <typename X> ostream& operator << (ostream& x,const vector<X>& v){for(int i=0;i<(int)v.size();++i) x<<v[i]<<" ";return x;} 
template <typename X> ostream& operator << (ostream& x,const set<X>& v){for(auto it:v) x<<it<<" ";return x;} 
template <typename X> ostream& operator << (ostream& x,const multiset<X>& v){for(auto it:v) x<<it<<" ";return x;} 
template <typename X, typename Y> ostream& operator << (ostream& x, const pair<X,Y>& v) {x<<v.ff<<" "<<v.ss;return x;} 
template <typename T, typename S> ostream& operator << (ostream& os, const map<T, S>& v) { for (auto it : v) os << it.first << "=>" << it.second << endl; return os; }
typedef tree <int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
struct pair_hash {inline std::size_t operator()(const std::pair<int, int> & v) const {return v.first*31+v.second;}};
const int MOD = 1e9 + 7;

int takemod(int a, int mod = MOD){a%=mod;if(a<0)a+=mod;return a;}

int fast_exp(int base, int expo, int mod = MOD) {
	int res=1;
	while(expo>0) {
		if(expo&1) res=(res*base)%mod;
		base=(base*base)%mod;
		expo>>=1;}
	return res;}

int modinv(int a, int mod = MOD){return takemod(fast_exp(takemod(a, mod), mod-2, mod), mod);}


void solve()
{
	int n;
	cin >> n;
	int a[n + 1]{};
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	map<int, int> m;
	int ans = 0;
	for(int i = 1; i <= n; i++) {
		int r = i - a[i];
		ans += m[r];
		m[i + a[i]]++;
	}
	cout << ans;
}

signed main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	#ifdef MADHAV
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("error.txt", "w", stderr);
	#endif
	clock_t t1, t2;
	t1 = clock();

	int tt = 1;
	// cin >> tt;
	while(tt--)
	{
		solve();
	}
	#ifdef MADHAV
	t2 = clock(); cout << "\n\ntime taken: " << (t2-t1)/(dbl)CLOCKS_PER_SEC;
	#endif
	return 0;
}