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.
#include <bits/stdc++.h>
using namespace std;
long long a[1000000], n,
    iCount = 0,
    pow2[38] = {0,           2,           4,           8,          16,
                32,          64,          128,         256,        512,
                1024,        2048,        4096,        8192,       16384,
                32768,       65536,       131072,      262144,     524288,
                1048576,     2097152,     4194304,     8388608,    16777216,
                33554432,    67108864,    134217728,   268435456,  536870912,
                1073741824,  2147483648,  4294967296,  8589934592, 17179869184,
                34359738368, 68719476736, 137438953472},
    aSub;
map<long long, long long> aMap;
int main() {
  scanf("%I64d", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
  }
  for (long long i = 1; i <= n; i++) {
    aMap[a[i]]++;
  }
  for (long long i = 1; i < n; i++) {
    aSub = 0;
    aMap[a[i]]--;
    for (int j = 1; j < 38; j++) {
      if (pow2[j] > a[i]) {
        aSub = pow2[j] - a[i];
        if (aMap[aSub] != NULL) {
          iCount += aMap[aSub];
        } else {
        }
      }
    }
  }
  printf("%I64d", iCount);
}
