One day, ZS the Coder wrote down an array of integers a with elements a1, a2, ..., an.

A subarray of the array a is a sequence al, al + 1, ..., ar for some integers (l, r) such that 1 ≤ l ≤ r ≤ n. ZS the Coder thinks that a subarray of a is beautiful if the bitwise xor of all the elements in the subarray is at least k.

Help ZS the Coder find the number of beautiful subarrays of a!

Input

The first line contains two integers n and k (1 ≤ n ≤ 106, 1 ≤ k ≤ 109) — the number of elements in the array a and the value of the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 109) — the elements of the array a.

Output

Print the only integer c — the number of beautiful subarrays of the array a.

Examples

Input

3 1
1 2 3


Output

5


Input

3 2
1 2 3


Output

3


Input

3 3
1 2 3


Output

2
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
const long long M = (long long)1e9 + 7;
int Int() {
  int x;
  scanf("%d", &x);
  return x;
}
long long Long() {
  long long x;
  scanf("%lld", &x);
  return x;
}
void Char(char *ch) { scanf("%s", ch); }
template <typename t>
inline t abs(t a) {
  if (a >= 0) return a;
  return -a;
}
template <typename T>
inline T modpow(T b, T p, T Mod) {
  T res = 1;
  while (p) {
    if (p & 1) {
      res *= b;
      res %= Mod;
    }
    b *= b;
    b %= Mod;
    p >>= 1;
  }
  return res;
}
template <typename T>
inline T ModInv(T b, T Mod) {
  return modpow(b, Mod - 2, Mod);
}
template <typename T>
inline T pwr(T b, T p) {
  T res = 1;
  while (p) {
    if (p & 1) {
      res *= b;
    }
    b *= b;
    p >>= 1;
  }
  return res;
}
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
inline void Int(T &n) {
  n = 0;
  int f = 1;
  register int ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
  n = n * f;
}
template <typename T, typename U>
inline void Int(T &a, U &b) {
  Int(a);
  Int(b);
}
template <typename T, typename U, typename W>
inline void Int(T &a, U &b, W &c) {
  Int(a, b);
  Int(c);
}
template <typename T, typename U, typename W, typename X>
inline void Int(T &a, U &b, W &c, X &d) {
  Int(a, b, c);
  Int(d);
}
vector<int> Unique(vector<int> v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  return v;
}
vector<string> split(const string &s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cout << it->substr((*it)[0] == ' ', it->length()) << " = " << a << " ";
  err(++it, args...);
}
struct node {
  int cnt;
  node *tree[2];
  node() {
    cnt = 0;
    tree[0] = tree[1] = NULL;
  }
} * root;
void add(int x) {
  node *cur = root;
  for (int i = 30; i >= 0; i--) {
    int f = (bool)(x & (1 << i));
    if (cur->tree[f] == NULL) {
      cur->tree[f] = new node();
    }
    cur = cur->tree[f];
    cur->cnt++;
  }
}
int sum(int x, int k) {
  int res = 0;
  node *cur = root;
  for (int i = 30; i >= 0; i--) {
    int p = (bool)(x & (1 << i));
    int q = (bool)(k & (1 << i));
    if (q == 0) {
      if (p == 1) {
        if (cur->tree[1] != NULL) {
          cur = cur->tree[1];
        } else {
          return res;
        }
      } else {
        if (cur->tree[0] != NULL) {
          cur = cur->tree[0];
        } else {
          return res;
        }
      }
    } else {
      if (p == 1) {
        if (cur->tree[1] != NULL) res += cur->tree[1]->cnt;
        if (cur->tree[0] != NULL) {
          cur = cur->tree[0];
        } else {
          return res;
        }
      } else {
        if (cur->tree[0] != NULL) res += cur->tree[0]->cnt;
        if (cur->tree[1] != NULL) {
          cur = cur->tree[1];
        } else {
          return res;
        }
      }
    }
  }
  return res;
}
int a[N];
int main() {
  int t = 1, tc = 0;
  while (t--) {
    root = new node();
    int n = Int(), k = Int();
    long long res = 0;
    add(0);
    for (int i = 1; i <= n; i++) {
      a[i] = Int();
      a[i] ^= a[i - 1];
      res += sum(a[i], k);
      add(a[i]);
    }
    long long s = (1LL * n * (n + 1LL)) / 2LL;
    printf("%lld\n", s - res);
  }
  return 0;
}
