Codeforces-959F：Mahmoud and Ehab and yet another xor task && CDOJ-2057：简单异或（线性基）

F. Mahmoud and Ehab and yet another xor task

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

Ehab has an array a of n integers. He likes the bitwise-xor operation 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

Copy

5 5
0 1 2 3 4
4 3
2 0
3 7
5 7
5 8

output

Copy

4
2
0
4
0

input

Copy

3 2
1 1 1
3 1
2 0

output

Copy

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.

思路：线性基。求出数组a的前缀线性基，那么对于询问l，x。判断x是否能由l处的前缀线性基表示，若不能则输出0；若能，则对于线性基里的数来说只有一种选法，然后非线性基里的数可选可不选，就算选了也可利用线性基将其异或补成x。

#include
using namespace std;
const int MAX=1e5+10;
const int MOD=1e9+7;
typedef long long ll;
int s[MAX][21],num[MAX];
int POW(int n)
{
    ll res=1,a=2;
    while(n)
    {
        if(n&1)res=res*a%MOD;
        a=a*a%MOD;
        n/=2;
    }
    return res;
}
int check(int x,int y)
{
    for(int i=20;i>=0;i--)
    {
        if(y&(1<=0;j--)
    {
        if(x&(1<>n>>q;
    memset(s,0,sizeof s);
    memset(num,0,sizeof num);
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<=20;j++)s[i][j]=s[i-1][j];
        int x;
        scanf("%d",&x);
        num[i]=add(s[i],x);
        num[i]+=num[i-1];
    }
    while(q--)
    {
        int l,x;
        scanf("%d%d",&l,&x);
        if(check(l,x)==0)puts("0");
        else printf("%d\n",POW(l-num[l]));
    }
    return 0;
}