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 PyPy3 (7.3.0) (judge verdict: Accepted)

def examA():
    S = SI()
    if S=="ARC":
        ans = "ABC"
    else:
        ans = "ARC"
    print(ans)
    return

def examB():
    N, K = LI()
    snuke = set(range(1,N+1))
    for _ in range(K):
        d = I()
        A = LI()
        for a in A:
            snuke.discard(a)
    ans = len(snuke)
    print(ans)
    return

def examC():
    N, M = LI()
    H = LI()
    good = set(range(1,N+1))
    for _ in range(M):
        a, b = LI()
        if H[a-1]>=H[b-1]:
            good.discard(b)
        if H[b-1]>=H[a-1]:
            good.discard(a)
    ans = len(good)
    print(ans)
    return

def examD():
    X = I()
    #print(1000**5-999**5)
    for a in range(-1000,1000):
        for b in range(-1000,1000):
            if a**5-b**5==X:
                print(a,b)
                return
    return

def examE():
    N = I()
    A = LI()
    B = [0]*N
    D = defaultdict(int)
    for i in range(N):
        B[i] = A[i]-i
        c = A[i]+i
        D[c] += 1
    #print(B)
    #print(D)
    ans = 0
    for b in B:
        ans += D[-b]
    print(ans)
    return

def examF():
    N, A, B, C = LI()
    S = [SI()for _ in range(N)]
    ans = [""]*N


    print(ans)
    return

import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(readline())
def LI(): return list(map(int,readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 10**(-12)
alphabet = [chr(ord('a') + i) for i in range(26)]

sys.setrecursionlimit(10**7)

if __name__ == '__main__':
    examE()

"""
142
12 9 1445 0 1
asd dfg hj o o
aidn
"""