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 long N = 1 << 20;
const long long Mod = 1e9 + 7;
struct Query {
  long long id, l, x;
  Query() {}
  Query(long long _id, long long _l, long long _x) { id = _id, l = _l, x = _x; }
  bool operator<(const Query& Q2) const { return l < Q2.l; }
};
long long a[N];
Query Q[N];
long long Ans[N];
set<long long> st1;
set<long long> st2;
vector<long long> tmp;
int main() {
  long long n, q;
  cin >> n >> q;
  for (long long i = 0; i < n; i++) scanf("%I64d", a + i);
  for (long long i = 0; i < q; i++) {
    long long l, x;
    scanf("%I64d%I64d", &l, &x);
    Q[i] = Query(i, --l, x);
  }
  sort(Q, Q + q);
  st1.insert(0);
  for (long long i = 1; i < N; i++) st2.insert(i);
  long long cur = 1;
  long long j = 0;
  for (long long i = 0; i < n; i++) {
    long long x = *st1.begin();
    long long y = a[i] ^ x;
    if (st1.count(y)) {
      cur *= 2;
      cur %= Mod;
    } else {
      tmp.clear();
      for (long long z : st1)
        assert(st2.count(z ^ a[i])), tmp.push_back(z ^ a[i]);
      for (long long z : tmp) st1.insert(z), st2.erase(z);
    }
    while (j < q && Q[j].l == i)
      Ans[Q[j].id] = (st1.count(Q[j].x) ? cur : 0), j++;
  }
  for (long long i = 0; i < q; i++) printf("%I64d\n", Ans[i]);
  if (0) assert(0);
}
