Ehab has an array a of n integers. He likes the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) and he likes to bother Mahmoud so he came up with a problem. He gave Mahmoud q queries. In each of them, he gave Mahmoud 2 integers l and x, and asked him to find the number of subsequences of the first l elements of the array such that their bitwise-xor sum is x. Can you help Mahmoud answer the queries?

A subsequence can contain elements that are not neighboring.

Input

The first line contains integers n and q (1 ≤ n, q ≤ 105), the number of elements in the array and the number of queries.

The next line contains n integers a1, a2, ..., an (0 ≤ ai < 220), the elements of the array.

The next q lines, each contains integers l and x (1 ≤ l ≤ n, 0 ≤ x < 220), representing the queries.

Output

For each query, output its answer modulo 109 + 7 in a newline.

Examples

Input

5 5
0 1 2 3 4
4 3
2 0
3 7
5 7
5 8


Output

4
2
0
4
0


Input

3 2
1 1 1
3 1
2 0


Output

4
2

Note

The bitwise-xor sum of the empty set is 0 and the bitwise-xor sum of a set containing one element is that element itself.
#include <bits/stdc++.h>
using namespace std;
const long long modn = 1000000007;
inline long long mod(long long x) { return x % modn; }
template <class num>
inline void rd(num &x) {
  char c;
  while (isspace(c = getchar()))
    ;
  bool neg = false;
  if (!isdigit(c))
    neg = (c == '-'), x = 0;
  else
    x = c - '0';
  while (isdigit(c = getchar())) x = (x << 3) + (x << 1) + c - '0';
  if (neg) x = -x;
}
int n;
const int N = 1 << 20;
int a[N];
int mrk[N], tot = 0;
int p2[N];
vector<pair<int, int> > qs[N];
int ans[N];
int main() {
  int i;
  int q;
  rd(n);
  rd(q);
  p2[0] = 1;
  for (i = 1; i < N; i++) p2[i] = mod(2ll * p2[i - 1]);
  for (i = 0; i < n; i++) rd(a[i]);
  for (i = 0; i < q; i++) {
    int l, x;
    rd(l);
    rd(x);
    qs[l].push_back(pair<int, int>(i, x));
  }
  vector<int> all;
  all.push_back(0);
  mrk[0] = 1;
  tot = 0;
  for (i = 0; i < n; i++) {
    if (!mrk[a[i]]) {
      vector<int> y;
      for (int x : all) mrk[a[i] ^ x] = 1, y.push_back(a[i] ^ x);
      for (int yy : y) all.push_back(yy);
      tot++;
    }
    for (pair<int, int> p : qs[i + 1]) {
      if (mrk[p.second]) ans[p.first] = p2[i + 1 - tot];
    }
  }
  for (i = 0; i < q; i++) printf("%d\n", ans[i]);
}
