Codeforces Round 383 (Div. 2) |
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Finished |

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brute force

math

number theory

*1500

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B. Arpa’s obvious problem and Mehrdad’s terrible solution

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputThere are some beautiful girls in Arpa’s land as mentioned before.

Once Arpa came up with an obvious problem:

Given an array and a number *x*, count the number of pairs of indices *i*, *j* (1 ≤ *i* < *j* ≤ *n*) such that , where is bitwise xor operation (see notes for explanation).

Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.

Input

First line contains two integers *n* and *x* (1 ≤ *n* ≤ 10^{5}, 0 ≤ *x* ≤ 10^{5}) — the number of elements in the array and the integer *x*.

Second line contains *n* integers *a*_{1}, *a*_{2}, ..., *a*_{n} (1 ≤ *a*_{i} ≤ 10^{5}) — the elements of the array.

Output

Print a single integer: the answer to the problem.

Examples

Input

2 3

1 2

Output

1

Input

6 1

5 1 2 3 4 1

Output

2

Note

In the first sample there is only one pair of *i* = 1 and *j* = 2. so the answer is 1.

In the second sample the only two pairs are *i* = 3, *j* = 4 (since ) and *i* = 1, *j* = 5 (since ).

A bitwise xor takes two bit integers of equal length and performs the logical xor operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. You can read more about bitwise xor operation here: https://en.wikipedia.org/wiki/Bitwise_operation#XOR.

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

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