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思路：
如果k不为n-1的话，就把k和n-1配，0和n-1-k配，其他相加为n-1就配一对
如果为n-1的话就把后四组和前四组单独配一下，其他相加为n-1配一对,尽量不要打乱中间的配对
Code
#include <bits/stdc++.h> // #define DEBUG freopen("_in.txt", "r", stdin); #
define DEBUG freopen("_in.txt", "r", stdin), freopen("_out.txt", "w", stdout);
typedef long long ll; using namespace std; const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll maxn = 1e6 + 10; const ll maxm = 1e7 + 10; const ll mod = 1e9 + 7;
const double pi = acos(-1); const double eps = 1e-8; ll T, n, k; ll arr[maxn],
brr[maxn]; int main() { // DEBUG; // printf("%lf",pow(2,16)); scanf("%lld", &T);
while (T--) { scanf("%lld%lld", &n, &k); if (k == n - 1) { if (n == 4) printf(
"-1\n"); else { printf("%lld %lld\n", n-2, n - 1); printf("0 2\n"); printf("1
%lld\n", n-3); printf("3 %lld\n", n-4); for (ll i = 4; i < n / 2; i++) { printf(
"%lld %lld\n", i, n - 1 - i); } } continue; } printf("%lld %lld\n", k, n - 1);
if (k != 0) printf("0 %lld\n", n - 1 - k); for (ll i = 0; i < n / 2; i++) { if (
i!= k && i != 0 && i != n - 1 - k && i != n - 1 && n - 1 - i != k && n - 1 - i
!= 0 && n - 1 - i != n - 1 - k && n - 1 - i != n - 1) { printf("%lld %lld\n", i,
n- 1 - i); } } // printf("\n"); } return 0; }
