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)

#include <bits/stdc++.h>
#if ONLINE_JUDGE
    #define DEBUG false
#else
    #define DEBUG true
#endif
#define REP(i, k, n) for (int i = (int)(k); i < (int)(n); i++)
#define rep(i, n) REP(i, 0, n)
#define rep1(i, n) REP(i, 1, n)
#define rREP(i, k, n) for (int i = (int)(n)-1; i >= k; i--)
#define rrep(i, n) rREP(i, 0, n)
#define all(x) (x).begin(),(x).end()
#define SORT(x) std::sort(all(x))
#define REV(x) std::reverse(all(x))
#define UNIQUE(x) x.erase(std::unique(all(x)), x.end());
#define MAX(x) *std::max_element(all(x))
#define MIN(x) *std::min_element(all(x))
#define LOWitr(x, n) std::lower_bound(all(x), n)
#define UPRitr(x, n) std::upper_bound(all(x), n)
#define cii(x) int x; std::cin >> x
#define cill(x) signed long long int x; std::cin >> x
#define cis(x) std::string x; std::cin >> x
#define co(x) std::cout << x << "\n"
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define truecheck assert
#define dump(x) if(DEBUG) std::cout<<#x<<" : "<<(x)<<"\n"
using LL = signed long long int;
template<class T>using V = std::vector<T>;
template<class T>using VV = std::vector<std::vector<T>>;
template<class T>using Q = std::queue<T>;
template<class T>using PQ = std::priority_queue<T>;
template<class T,class U>using P = std::pair<T, U>;
template<class T,class U>using M = std::map<T, U>;
template<class T,class U>using UM = std::unordered_map<T, U>;
template<class T,class U>inline bool chmax(T &a, const U &b){if (a<b){a=b;return true;}return false;}
template<class T,class U>inline bool chmin(T &a, const U &b){if (b<a){a=b;return true;}return false;}
template<class T,class U>inline void vcinpb(T &v, U n){rep(i, n){typename T::value_type a;std::cin >> a;v.pb(a);}}
template<class T>inline void dumpv(const T& v, std::ostream& os = std::cout){
    if(DEBUG){
        os<<"{";
        auto f=[&os](typename T::value_type p){os<<p<<',';};
        std::for_each(all(v),f);
        os<<"}\n";
    }
}
const int INF = INT_MAX, MOD = 1e9 + 7;
const LL LLINF = LLONG_MAX;
// M_PI 3.1415...

void Main(){
    using std::cout;
    using std::cin;
    cill(n);
    V<LL> v(n), ph(201000), mh(201000), idxes;
    LL ans = 0;
    for(auto&& i : v) cin >> i;
    rep(i, n){
        if(i+1 + v[i] < 200900){
            ph[ i+1 + v[i] ]++;
            if (ph[ i+1 + v[i] ] == 1){
                idxes.pb(i+1 + v[i]);
            }
        }
        if(i+1 - v[i] > 0){
            mh[ i+1 - v[i] ]++;
            if (mh[ i+1 - v[i] ] == 1){
                idxes.pb(i+1 - v[i]);
            }
        }

    }
    SORT(idxes);
    UNIQUE(idxes);
    dumpv(idxes);
    for(auto&& i : idxes){
        ans += ph[i] * mh[i];
        dump(ans);
    }
    co(ans);
}

int main() {
    std::cin.tie(nullptr);std::ios::sync_with_stdio(false);
    std::cout << std::fixed << std::setprecision(15);
    Main();
    return 0;
}