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description:

题意有点难读 具体化的讲就是有n个点，每个点有点权。用这n个点来构造q个无向图，对于每个无向图Ei来说
规定属于Ei的点x,y. 满足Valx ^ Valy = k 就会有一条权值为1的边，问点x到点y的最短距离

solution:

首先是异或的一个性质 a ^ b = c <-> a ^ c = b <-> b ^ c = a 这三个是等价的
1.当a[x] ^ a[y] == k 时 他们之间有条边 输出1即可
2.当a[x] == a[y] 的时候 通过一个中间点来到达 即 k ^ c 判断是否存在 走了2步
3. 无法到达
code:

#include <bits/stdc++.h>

using namespace std;

#define LL long long
#define ULL unsigned long long
#define mes(x, a) memset(x, a, sizeof(x));
#define sca(a) scanf("%d", &a)
#define lowbit(x) x &(-x)
#define mk make_pair
#define pb(x) push_back(x)
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define lson v << 1
#define rson v << 1 | 1
#define pii pair<int, int>

inline void read(int &x) {
  x = 0;
  int flag_read = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag_read = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  x *= flag_read;
}

void out(int x) {
  if (x > 9) {
    out(x / 10);
  }
  putchar(x % 10 + '0');
}

const int N = 1e7 + 50;
int a[N];
int vis[N];

int main() {
  int n, q;
  scanf("%d%d",&n,&q);

  for (int i = 1; i <= n; i++) {
    scanf("%d",&a[i]);
    vis[a[i]] = 1;
  }

  while (q--) {
    int k, x, y;
    scanf("%d%d%d",&k,&x,&y);

    x = a[x], y = a[y];

    if ((x ^ y) == k) {
      puts("1");
    } else if (x == y && vis[k ^ x]) {
      puts("2");
    } else {
      puts("-1");
    }
  }
  return 0;
}