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)

/*!enjoy karo yaar!*/

//CF,CC,AtC,SPOJ: hp1999
//HE: hemant269
//HR: hemant2132

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

#define     int             long long int
#define     fast()          ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define     all(x)          (x).begin(),(x).end()
#define     rz(x)           resize(x)
#define     mem(a,b)        memset(a,b,sizeof(a))
#define     sz(x)           ((int)(x.size()))
#define     eb              emplace_back
#define     pb              push_back
#define     pf              push_front
#define     pob             pop_back
#define     pof             pop_front
#define     ins             insert
#define     vi              vector<int>
#define     pii             pair<int,int>
#define     mii             map<int,int>
#define     F               first
#define     S               second
#define     makep           make_pair
#define     maket           make_tuple
#define     remax(a,b)      a=max(a,b)
#define     remin(a,b)      a=min(a,b)
#define     bitcount(x)     __builtin_popcountll(x)
#define     iceil(n,x)      (((n)+(x)-1)/(x))
#define     gcd(a,b)        (__gcd((a),(b)))
#define     lcm(a,b)        (((a)*(b))/gcd((a),(b)))
#define     dbug(x)         cout<<#x<<": "<<(x)<<"\n"
#define     flush           fflush(stdout)
#define     show(x)         for(auto zz:x)cout<<zz<<" ";cout<<"\n";
#define     show2(x)        for(auto zz:x)cout<<zz.F<<" "<<zz.S<<"\n";

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;

const ld pi=acos(-1);
const ll inf=1e18,M=1e9+7;
const int N=1;

void solve()
{
    int n;
    cin>>n;

    vi v(n);
    for(int i=0;i<n;++i)
        cin>>v[i];

    vi a(n);
    for(int i=0;i<n;++i)
        a[i]=(i+1)-v[i];

    mii mp;

    int ans=0;
    for(int i=0;i<n;++i)
    {
        ans+=mp[a[i]];

        ++mp[(i+1)+v[i]];
    }

    cout<<ans;
}

int32_t main()
{
    fast();

    int t=1;
    //cin>>t;
    for(int z=1;z<=t;++z)
    {
        solve();
        //cout<<"\n";
    }

    return 0;
}