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;
int lnb[50], n, a[101000], q, l, x, usd, nus;
int sub[101000], pw[101050];
const int Md = 1e9 + 7;
struct qur {
  int l, x, t;
} qr[100100];
bool cmp(qur p, qur q) {
  if (p.l != q.l)
    return p.l < q.l;
  else
    return p.x < q.x;
}
void gtlnb(int k) {
  for (int i = 30; i >= 0; i--) {
    if ((k >> i) & 1) {
      if (!lnb[i]) {
        lnb[i] = k;
        usd++;
        break;
      }
      k ^= lnb[i];
    }
  }
}
bool cxor(int k) {
  for (int i = 30; i >= 0; i--) {
    if ((k >> i) & 1) {
      if (lnb[i])
        k ^= lnb[i];
      else
        return false;
    }
  }
  return true;
}
void pw2() {
  int k = 1;
  for (int i = 0; i <= 101010; i++) {
    pw[i] = k % Md;
    k *= 2;
    k %= Md;
  }
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &qr[i].l, &qr[i].x);
    qr[i].t = i;
  }
  pw2();
  sort(qr + 1, qr + q + 1, cmp);
  for (int i = 1; i <= q; i++) {
    while (nus < qr[i].l) gtlnb(a[++nus]);
    if (cxor(qr[i].x)) sub[qr[i].t] = pw[nus - usd];
  }
  for (int i = 1; i <= q; i++) printf("%d\n", sub[i]);
}
