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 Python (3.8.2) (judge verdict: Accepted)

import sys
sys.setrecursionlimit(1000000000)
import math
from math import gcd
def lcm(a, b): return a * b // gcd(a, b)
from itertools import count, permutations, chain
from functools import lru_cache
from collections import deque, defaultdict
from pprint import pprint
ii = lambda: int(input())
mis = lambda: map(int, input().split())
lmis = lambda: list(mis())
INF = float('inf')
N1097 = 10**9 + 7

def meg(f, ok, ng):
    while abs(ok-ng)>1:
        mid = (ok+ng)//2
        if f(mid):
            ok=mid
        else:
            ng=mid
    return ok

def get_inv(n, modp):
    return pow(n, modp-2, modp)

def factorials_list(n, modp):    # 10**6
    fs = [1]
    for i in range(1, n+1):
        fs.append(fs[-1] * i % modp)
    return fs

def invs_list(n, fs, modp):     # 10**6
    invs = [get_inv(fs[-1], modp)]
    for i in range(n, 1-1, -1):
        invs.append(invs[-1] * i % modp)
    invs.reverse()
    return invs

def comb(n, k, modp):
    num = 1
    for i in range(n, n-k, -1):
        num = num * i % modp
    den = 1
    for i in range(2, k+1):
        den = den * i % modp
    return num * get_inv(den, modp) % modp

def comb_from_list(n, k, modp, fs, invs):   
    return fs[n] * invs[n-k] * invs[k] % modp

#

class UnionFindEx:
    def __init__(self, size):
        #正なら根の番号、負ならグループサイズ
        self.roots = [-1] * size
    def getRootID(self, i):
        r = self.roots[i]
        if r < 0:   #負なら根
            return i
        else:
            r = self.getRootID(r)
            self.roots[i] = r
            return r
    def getGroupSize(self, i):
        return -self.roots[self.getRootID(i)]
    def connect(self, i, j):
        r1, r2 = self.getRootID(i), self.getRootID(j)
        if r1 == r2:
            return False
        if self.getGroupSize(r1) < self.getGroupSize(r2):
            r1, r2 = r2, r1
        self.roots[r1] += self.roots[r2]    #サイズ更新
        self.roots[r2] = r1
        return True

Yes = 'Yes'
No = 'No'


def main():
    N=ii()
    A=[None] + lmis()
    d = {}
    for i in range(1, N+1):
        A[i] -= i
        key = A[i]
        if key in d:
            d[key] += 1
        else:
            d[key] = 1
    ans = 0
    for i in range(1, N+1):
        b = -(A[i]+2*i)
        if b in d:
            ans += d[b]
    print(ans)




main()