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 int maxn = 1e5 + 5, mod = 1e9 + 7;
int n, Q, a[maxn], siz, pw[maxn], ans[maxn];
struct question {
  int l, x, bh;
  inline bool operator<(const question &rhs) const { return l < rhs.l; }
} q[maxn];
struct XXJ {
  int a[21];
  inline void insert(int x) {
    for (int i = 20; i >= 0; --i) {
      if (!((x >> i) & 1))
        continue;
      else {
        if (a[i])
          x ^= a[i];
        else {
          a[i] = x;
          siz++;
          return;
        }
      }
    }
  }
  inline int query(int sum, int x) {
    for (int i = 20; i >= 0; --i) {
      if (a[i] && ((x >> i) & 1)) x ^= a[i];
    }
    if (x == 0)
      return pw[sum - siz];
    else
      return 0;
  }
} xxj;
int main() {
  scanf("%d", &n);
  scanf("%d", &Q);
  pw[0] = 1;
  for (int i = 1; i <= n; ++i)
    scanf("%d", &a[i]), pw[i] = (pw[i - 1] * 2) % mod;
  for (int i = 1; i <= Q; ++i) {
    scanf("%d%d", &q[i].l, &q[i].x);
    q[i].bh = i;
  }
  sort(q + 1, q + Q + 1);
  int last = 0;
  for (int i = 1; i <= Q; ++i) {
    for (int j = last + 1; j <= q[i].l; ++j) {
      xxj.insert(a[j]);
      last = j;
    }
    ans[q[i].bh] = xxj.query(q[i].l, q[i].x);
  }
  for (int i = 1; i <= Q; ++i) printf("%d\n", ans[i]);
  return 0;
}
