You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2 (i. e. some integer x exists so that ai + aj = 2x).

Input

The first line contains the single positive integer n (1 ≤ n ≤ 105) — the number of integers.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2.

Examples

Input

4
7 3 2 1


Output

2


Input

3
1 1 1


Output

3

Note

In the first example the following pairs of indexes include in answer: (1, 4) and (2, 4).

In the second example all pairs of indexes (i, j) (where i < j) include in answer.
from sys import stdin, stdout
import math
n = int(stdin.readline()[:-1])
a = sorted([int(num) for num in stdin.readline()[:-1].split(' ')])
b = [(2** (math.floor(math.log(num, 2)) + 1) - num) for num in a]
ans = 0
'''
    in pratica ci metto la coppia k:x,y tale che
    k è b[i]
    x è il numero di volte che si trova prima di y
'''
d = {}
da = {num: [0, 0] for num in a}
for i in range(n):
    da[a[i]][0] += 1
    da[a[i]][1] = b[i]
for i in da:
    if i == da[i][1]:
        ans += da[i][0]*(da[i][0] - 1)//2

    elif da[i][1] in da:
        ans += da[da[i][1]][0] * da[i][0]
'''
for i in range(n):
    if b[i] not in d:
        d[b[i]] = {'x': a[:i].count(b[i]), 'y': i}
    else:
        d[b[i]]['x'] += a[d[b[i]]['y']:i].count(b[i])
        d[b[i]]['y'] = i
    ans += d[b[i]]['x']
'''
stdout.write(str(ans))
