// APPROX2, Setter's solution

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

#define maxn 100000 + 5

int n, m, k, i, j, a[maxn], opt, ans, tn;

int main (int argc, char * const argv[]) {
	scanf("%d", &tn);
	for(; tn; tn--) {
		scanf("%d %d", &n, &k);
		for(i = 1; i <= n; i++) scanf("%d", &a[i]); // reading input
		opt = 2000000000;
		ans = 0;
		for(i = 1; i <= n; i++) for(j = i + 1; j <= n; j++) if (abs(a[i] + a[j] - k) < opt) { // finding the closest pair
			opt = abs(a[i] + a[j] - k);
			ans = 1;
		} else if (abs(a[i] + a[j] - k) == opt) ++ans; // counting the number of the closest pairs
		printf("%d %d\n", opt, ans);
	}
    return 0;
}