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;
using piii = pair<pair<int, int>, int>;
constexpr int maxn = 1e5 + 10;
constexpr int mod = 1e9 + 7;
int a[maxn], resp[maxn];
vector<int> x;
int get(int v) {
  for (auto it : x) v = min(v, v ^ it);
  return v;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  vector<piii> query(q);
  int cnt = 0;
  for (auto& it : query)
    scanf("%d %d", &it.first.first, &it.first.second), it.second = cnt++;
  sort(query.begin(), query.end());
  int ans = 1;
  for (int i = 1, ptr = 0; i <= n; i++) {
    if (get(a[i]) == 0)
      ans *= 2, ans %= mod;
    else
      x.push_back(get(a[i]));
    for (; ptr < q && query[ptr].first.first == i; ptr++)
      if (get(query[ptr].first.second) == 0)
        resp[query[ptr].second] = ans;
      else
        resp[query[ptr].second] = 0;
  }
  for (int i = 0; i < q; i++) printf("%d\n", resp[i]);
}
