There 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 <image>, where <image> is bitwise xor operation (see notes for explanation).

<image>

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 ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the array and the integer x.

Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — 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. <image> so the answer is 1.

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

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>.
import sys

########################################################################
_TOKENS = []
_TOKENS_POS = 0
def gs():
    global _TOKENS, _TOKENS_POS
    while _TOKENS_POS == len(_TOKENS):
        _TOKENS = raw_input().split(" ")
        _TOKENS_POS = 0
        while _TOKENS_POS < len(_TOKENS) and _TOKENS[_TOKENS_POS] == "":
            _TOKENS_POS += 1
    _TOKENS_POS += 1
    return _TOKENS[_TOKENS_POS-1]
def gi():
    return int(gs())
def pr(a, newline=True):
    if not isinstance(a, basestring):
        a = str(a)
    if newline:
        a += "\n"
    sys.stdout.write(a)
    sys.stdout.flush()

n = gi()
x = gi()

c = {}
for i in xrange(n):
    a = gi()
    c[a] = c.get(a, 0) + 1

r = 0
if x == 0:
    for v in c.itervalues():
        r += v*(v-1)/2
else:
    for k, v in c.iteritems():
        k2 = k ^ x
        if k2 > k:
            r += v * c.get(k2, 0)

print r
