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 = 63;
using T = int;
struct Gaussian {
  int totalBits;
  vector<T> v;
  Gaussian() {}
  Gaussian(int _totalBits) : totalBits(_totalBits) {}
  Gaussian(int _totalBits, vector<T> _v) : totalBits(_totalBits), v(_v) {}
  void insert(T val) { v.push_back(val); }
  void eliminate() {
    int row = 0;
    int n = (int)v.size();
    vector<int> where(totalBits + 1, -1);
    for (int bit = totalBits; row < n && bit >= 0; bit--) {
      bool check = false;
      for (int i = row; i < n; i++) {
        if ((v[i] >> bit) & 1) {
          swap(v[i], v[row]);
          check = true;
          break;
        }
      }
      if (!check) continue;
      where[bit] = row;
      for (int i = row + 1; i < n; i++) {
        if ((v[i] >> bit) & 1) {
          v[i] ^= v[row];
        }
      }
      row++;
    }
    while (!v.empty() && !v.back()) v.pop_back();
  }
  bool check(T val) {
    for (int& i : v) {
      for (int j = totalBits; j >= 0; j--) {
        if ((i >> j) & 1) {
          if ((val >> j) & 1) {
            val ^= i;
          }
          break;
        }
      }
    }
    return val != 0;
  }
  T findMax() {
    eliminate();
    T ans = v[0];
    for (int i = 1; i < (int)v.size(); i++) {
      if ((ans ^ v[i]) > ans) ans = ans ^ v[i];
    }
    return ans;
  }
};
const int MOD = (int)1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  vector<int> power(n + 1, 1);
  for (int i = 1; i <= n; ++i) {
    power[i] = (power[i - 1] * 2LL) % MOD;
  }
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  vector<vector<pair<int, int>>> Q(n);
  vector<int> ans(q, 0);
  for (int i = 0; i < q; i++) {
    int l, x;
    cin >> l >> x;
    l--;
    Q[l].push_back({x, i});
  }
  Gaussian gauss(21);
  for (int i = 0; i < n; i++) {
    gauss.insert(a[i]);
    gauss.eliminate();
    for (auto& j : Q[i]) {
      if (!gauss.check(j.first)) {
        ans[j.second] = power[i + 1 - gauss.v.size()];
      }
    }
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << " \n"[i == q - 1];
  }
  return 0;
}
