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 N = 1e5 + 10;
const long long mod = 1e9 + 7;
int n, q;
int a[N];
long long two[N];
int ans[N];
struct LB {
  int p[32];
  void init() { memset(p, 0, sizeof p); }
  bool insert(int x) {
    for (int i = 21; i >= 0; i--) {
      if ((x >> i) & 1) {
        if (p[i] == 0) {
          p[i] = x;
          return true;
        }
        x ^= p[i];
      }
    }
    return false;
  }
  int check(int x) {
    int res = 0;
    for (int i = 21; i >= 0; i--) {
      if ((x >> i) & 1) {
        if (p[i] == 0)
          return -1;
        else
          x ^= p[i];
        res++;
      }
    }
    return res;
  }
} lb;
struct Query {
  int l, id, x;
} b[N];
bool cmp(Query a, Query b) { return a.l < b.l; }
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", &b[i].l, &b[i].x);
    b[i].id = i;
  }
  two[0] = 1;
  for (int i = 1; i < N; i++) two[i] = two[i - 1] * 2 % mod;
  sort(b + 1, b + 1 + q, cmp);
  int l = 0, size = 0;
  for (int i = 1; i <= q; i++) {
    while (l < b[i].l) {
      l++;
      if (lb.insert(a[l])) size++;
    }
    int res = lb.check(b[i].x);
    if (res != -1)
      ans[b[i].id] = two[l - size];
    else
      ans[b[i].id] = 0;
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
