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 maxn = 1e5 + 10;
const int mod = 1e9 + 7;
int l[maxn], a[maxn], A[maxn];
map<pair<int, int>, int> Ans;
pair<int, int> p[maxn];
struct space {
  vector<int> val;
  bool add(int x) {
    for (auto i : val)
      if (x & i & -i) x ^= i;
    if (!x) return 0;
    for (int i = 0; i < val.size(); i++)
      if (val[i] & x & -x) val[i] ^= x;
    val.push_back(x);
    return 1;
  }
  bool has(int x) {
    for (auto i : val)
      if (x & i & -i) x ^= i;
    return (x == 0);
  }
} s;
int pw[maxn];
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0);
  pw[0] = 1;
  for (int i = 1; i < maxn; i++) pw[i] = (2 * pw[i - 1]) % mod;
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> A[i];
  for (int i = 0; i < q; i++) {
    cin >> l[i] >> a[i];
    p[i] = {l[i], a[i]};
  }
  sort(p, p + q);
  int cur = 0, sz = 0;
  for (int ii = 0; ii < q; ii++) {
    auto i = p[ii];
    int len = i.first, val = i.second;
    while (cur < len) sz += s.add(A[cur++]);
    if (s.has(val)) Ans[{len, val}] = pw[len - sz];
  }
  for (int i = 0; i < q; i++) cout << Ans[{l[i], a[i]}] << '\n';
}
