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;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
template <class Iter>
void rprintf(const char *fmt, Iter begin, Iter end) {
  for (bool sp = 0; begin != end; ++begin) {
    if (sp)
      putchar(' ');
    else
      sp = true;
    printf(fmt, *begin);
  }
  putchar('\n');
}
const long long MOD = 1e9 + 7;
long long powMod(long long x, long long y) {
  x %= MOD;
  long long r = 1;
  for (; y; y >>= 1) {
    if (y & 1) r = r * x % MOD;
    x = x * x % MOD;
  }
  return r;
}
int N, Q;
struct Query {
  int l, x, i;
  bool operator<(const Query &y) const { return l < y.l; }
} query[100011];
;
int A[100011];
int D[30];
int ans[100011];
void MAIN() {
  scanf("%d%d", &N, &Q);
  for (int i = 0, i_len = (N); i < i_len; ++i) scanf("%d", A + i);
  for (int i = 0, i_len = (Q); i < i_len; ++i) {
    scanf("%d%d", &query[i].l, &query[i].x);
    query[i].i = i;
  }
  sort(query, query + Q);
  int rk = 0;
  int cur = 0;
  for (int i = 0, i_len = (Q); i < i_len; ++i) {
    while (cur < query[i].l) {
      int a = A[cur++];
      for (int t = 20; t--;) {
        if (a >> t & 1) a ^= D[t];
        if (a >> t & 1) {
          rk++;
          D[t] = a;
          break;
        }
      }
    }
    {
      int a = query[i].x;
      for (int t = 20; t--;) {
        if (a >> t & 1) a ^= D[t];
      }
      if (a == 0) {
        ans[query[i].i] = powMod(2, cur - rk);
      } else {
        ans[query[i].i] = 0;
      }
    }
  }
  for (int i = 0, i_len = (Q); i < i_len; ++i) printf("%d\n", ans[i]);
}
int main() {
  int TC = 1;
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}
