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 = 1e9 + 7;
int n, q;
int a[100010];
long long Pow(long long foo, long long bar) {
  long long ans = 1ll;
  while (bar) {
    if (bar % 2) ans = (ans * foo) % mod;
    foo = (foo * foo) % mod;
    bar /= 2;
  }
  return ans;
}
struct TQuerry {
  int pos;
  int value;
  int id;
  long long ans;
  TQuerry() {}
  TQuerry(int foo, int bar, int magic) {
    pos = foo;
    value = bar;
    id = magic;
  }
  bool operator<(const TQuerry &op) const { return id < op.id; }
} Querry[100010];
bool cmp(TQuerry foo, TQuerry bar) { return foo.pos < bar.pos; }
bool vis[2000010];
vector<int> lis;
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= q; ++i) {
    int pos;
    int value;
    cin >> pos >> value;
    Querry[i] = TQuerry(pos, value, i);
  }
  sort(Querry + 1, Querry + q + 1, cmp);
  lis.push_back(0);
  vis[0] = true;
  int foo = 0, cur = 0;
  for (int i = 1; i <= n; ++i) {
    if (vis[a[i]])
      foo++;
    else {
      int bar = (int)lis.size();
      for (int j = 0; j < bar; ++j) {
        lis.push_back(a[i] ^ lis[j]);
        vis[a[i] ^ lis[j]] = true;
      }
    }
    while (cur < q && Querry[cur + 1].pos <= i) {
      cur++;
      if (!vis[Querry[cur].value])
        Querry[cur].ans = 0;
      else
        Querry[cur].ans = Pow(2, 1ll * foo);
    }
  }
  sort(Querry + 1, Querry + q + 1);
  for (int i = 1; i <= q; ++i) cout << Querry[i].ans << endl;
  return 0;
}
