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 INF = 1e18;
const long long mod = 1e9 + 7;
const int N = 1e5 + 10;
const int LOGN = 30;
int base[LOGN];
int arr[N];
int ans[N];
long long pows[N];
vector<pair<int, int> > queries[N];
int basis_size = 0;
bool try_gauss(int v, int more) {
  for (int i = LOGN - 1; i >= 0; i--) {
    if (base[i] and (v & (1 << i))) v ^= base[i];
  }
  if (v == 0) return true;
  if (!more) return false;
  for (int i = LOGN - 1; i >= 0; i--) {
    if (v & (1 << i)) {
      base[i] = v;
      basis_size++;
      return false;
    }
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, q;
  cin >> n >> q;
  pows[0] = 1;
  for (int i = 1; i <= n; i++) pows[i] = (2 * pows[i - 1]) % mod;
  memset(ans, 0, sizeof(ans));
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    queries[i].clear();
  }
  for (int i = 1; i <= q; i++) {
    int l, x;
    cin >> l >> x;
    queries[l].push_back(make_pair(i, x));
  }
  int xorval = 0;
  for (int i = 1; i <= n; i++) {
    xorval ^= arr[i];
    try_gauss(xorval, 1);
    for (int j = 0; j < queries[i].size(); j++) {
      if (try_gauss(queries[i][j].second, 0)) {
        ans[queries[i][j].first] = pows[i - basis_size];
      }
    }
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << endl;
  return 0;
}
