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>
void rd(T& ret) {
  ret = 0;
  bool ok = 0, u = 0;
  for (;;) {
    int c = getchar();
    if (c >= '0' && c <= '9')
      ret = (ret << 3) + (ret << 1) + c - '0', ok = 1;
    else if (c == '-')
      u = 1;
    else if (ok) {
      if (u) ret *= -1;
      return;
    }
  }
}
long long powmod(long long p, long long n, long long mod) {
  long long ret = 1;
  for (; n; n >>= 1) {
    if (n & 1) ret = ret * p % mod;
    p = p * p % mod;
  }
  return ret;
}
template <class T>
bool chmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool chmax(T& a, const T& b) {
  return b > a ? a = b, 1 : 0;
}
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
const long long mod = (long long)1e9 + 7;
const int N = 1 << 21;
pair<pair<int, int>, int> qn[N];
int stk[N], top, an[N], ans[N];
bool has[N];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = (0); i < (n); i++) scanf("%d", an + i);
  for (int i = (0); i < (q); i++) {
    int l, first;
    scanf("%d%d", &l, &first);
    --l;
    qn[i].first = pair<int, int>(l, first);
    qn[i].second = i;
  }
  sort(qn, qn + q);
  has[0] = 1;
  stk[top++] = 0;
  long long t = 1, it = 0;
  for (int i = (0); i < (n); i++) {
    if (has[an[i]])
      t = (t << 1) % mod;
    else {
      int tn = top, second;
      for (int j = (0); j < (tn); j++) {
        second = an[i] ^ stk[j];
        stk[top++] = second;
        has[second] = 1;
      }
      has[an[i]] = 1;
      stk[top++] = an[i];
    }
    while (it < q && qn[it].first.first == i) {
      int val = qn[it].first.second;
      ans[qn[it].second] = has[val] ? t : 0;
      ++it;
    }
  }
  for (int i = (0); i < (q); i++) printf("%d\n", ans[i]);
  return 0;
}
