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 mod = (int)1e9 + 7;
const int maxN = (int)1e5 + 10;
int pw[maxN];
int sum(int a, int b) {
  int s = a + b;
  if (s >= mod) s -= mod;
  return s;
}
int n, q;
vector<int> p[maxN];
int a[maxN];
bool used[maxN];
bool ok(vector<int> &s, int x) {
  if (s.size() == 0) return (x == 0);
  int l = 0;
  for (int i = 19; i >= 0; i--) {
    if ((x & (1 << i)) == 0) continue;
    int t = -1;
    for (; l < s.size(); l++) {
      if (s[l] & (1 << i)) {
        t = l;
        x ^= s[l];
        break;
      }
    }
    if (t == -1) return false;
  }
  return (x == 0);
}
void reduce(vector<int> &s) {
  memset(used, 0, sizeof used);
  for (int i = 19; i >= 0; i--) {
    int t = -1;
    for (int j = 0; j < s.size(); j++) {
      if (used[j]) continue;
      if (s[j] & (1 << i)) {
        t = j;
      }
    }
    if (t == -1) continue;
    used[t] = true;
    for (int j = 0; j < s.size(); j++) {
      if (j == t) continue;
      if (s[j] & (1 << i)) s[j] ^= s[t];
    }
  }
  sort(s.begin(), s.end());
  reverse(s.begin(), s.end());
}
int main() {
  pw[0] = 1;
  for (int i = 1; i < maxN; i++) pw[i] = sum(pw[i - 1], pw[i - 1]);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (ok(p[i - 1], a[i]))
      p[i] = p[i - 1];
    else {
      p[i] = p[i - 1];
      p[i].push_back(a[i]);
      reduce(p[i]);
    }
  }
  for (int i = 1; i <= q; i++) {
    int l, x;
    cin >> l >> x;
    if (ok(p[l], x))
      cout << pw[l - p[l].size()] << '\n';
    else
      cout << 0 << '\n';
  }
  return 0;
}
