A. And Matching

// Problem: A. And Matching
// Contest: Codeforces - Codeforces Round #768 (Div. 1)
// URL: https://codeforces.com/problemset/problem/1630/A
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-04-12 19:59:19
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a;
        a = a * a;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
#define int long long

int n, k;
int c (int x) {
    return x ^(n - 1);
}
void solve() {
    cin >> n >> k;
    vpll ans;
    map<int, int> st;
    if (k == 0) {
        for (int i = 0; i < n / 2; i++) {
            ans.push_back({i, c(i)});
        }
    }
    else if (k == n - 1) {
        if (n == 4) {
            puts("-1");
            return;
        }
        ans.push_back({n - 1, n - 2});
        ans.push_back({0, 2});
        ans.push_back({1, n - 3});
        st[n - 1] = st[n - 2] = st[0] = st[2]
        = st[1] = st[n - 3] = 1;
        for (int i = 0;ans.size() < n / 2; i ++) {
            if (st[i] || st[c(i)]) continue;
            ans.push_back({i, c(i)});
        }
    }
    else  {
        ans.pb({k, n - 1});
        ans.pb({0, c(k)});
        st[k] = st[n - 1] = st[0] = st[c(k)] = 1;
        for (int i = 0; ans.size() < n / 2; i ++) {
            if (st[i] || st[c(i)]) continue;
            
            ans.pb({i, c(i)});
        }
    }
    for (auto t : ans) {
        cout << t.fi << " " << t.se << endl;
    }
}
signed main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}

k = 0 我们可以构造c(i)&i 对于k==n - 1 对于n = 4 打表可以发现不可能等于3 n= 8 我们可以先构造一个 (n - 1) & (n - 2) = n - 2 1 * (n - 3) = 1; 0 & 2 = 0 其余用i&c(i) 我在想的时候陷入了对每个值进行考虑的困境，一致无法得到想要的结果 问题结构:分类讨论题，通过讨论边界的条件下的情况，可以为一般情况提供思路。0的运用，通过让一些项等于0，其他的只让某几个项得到我们想要的值