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 double PI = 3.1415926535897932384626433;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long N = 1e5 + 5, p = 1e9 + 7;
long long n, a[N], ans[N], q, rnk, pw[N];
vector<pair<long long, long long> > v[N];
vector<long long> basis;
void add(long long x) {
  for (long long bit = (20 - 1); bit >= 0; bit--) {
    if (x == 0) break;
    if ((x & (1LL << bit)) != 0) {
      if (basis[bit] == -1) {
        basis[bit] = x;
        rnk++;
        break;
      } else
        x ^= basis[bit];
    }
  }
}
bool chk(long long x) {
  for (long long bit = (20 - 1); bit >= 0; bit--) {
    if (x == 0) return 1;
    if ((x & (1LL << bit)) != 0) {
      if (basis[bit] == -1)
        return 0;
      else
        x ^= basis[bit];
    }
  }
  return 1;
}
void func() {
  cin >> n >> q;
  for (long long i = 1; i < n + 1; i++) cin >> a[i];
  for (long long i = 1; i < q + 1; i++) {
    long long l, x;
    cin >> l >> x;
    v[l].emplace_back(x, i);
  }
  pw[0] = 1;
  for (long long i = 1; i < N; i++) pw[i] = (pw[i - 1] * 2) % p;
  basis.resize(20, -1);
  for (long long i = 1; i < n + 1; i++) {
    add(a[i]);
    for (auto query : v[i]) {
      if (!chk(query.first))
        ans[query.second] = 0;
      else
        ans[query.second] = pw[i - rnk];
    }
  }
  for (long long i = 1; i < q + 1; i++) cout << ans[i] << "\n";
}
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  }
  long long ntc = 1;
  for (long long i = 1; i < ntc + 1; i++) {
    func();
  }
  return 0;
}
