/*
HDU 6076 - Security Check [ DP,二分 ] | 2017 Multi-University Training Contest 4
题意:
给出两个检票序列 A[N], B[N]
规定 abs(A[i]-B[j]) <= k 的i,j不能同时检票
求最少的检票时间
限制 N<= 6e4, k <= 10
分析:
f(i,j) 为检票至i,j的时间
则 f(i,j) = f(i-1,j-1) + 1 , abs(A[i]-B[j]) > k
= min(f(i-1,j), f(i,j-1)) + 1 , abs(A[i]-B[j]) <= k
对于第二项,由于k小,可DP
对于第一项,可以二分最大的 t 使得 f(i,j) = f(i-t,j-t) + t 成立
那么 f(i-t,j-t) 就是第二项的了
*/
#include <bits/stdc++.h>
using namespace std;
const int N = 60005;
vector<int> G[N<<1];
int dp[N][25];
int a[N], b[N], pos[N];
int t, n, k;
int f(int n, int m)
{
if (n == 0 || m == 0) return n+m;
if (abs(a[n]-b[m]) > k)
{
int t = lower_bound(G[m-n+N].begin(), G[m-n+N].end(), n) - G[m-n+N].begin();
if (t == 0) return max(n, m);
t = G[m-n+N][t-1];
return f(t, m-n+t) + n-t;
}
else
{
int t = b[m]-a[n]+k;
if (dp[m][t] == -1)
dp[m][t] = min(f(n-1, m), f(n, m-1)) + 1;
return dp[m][t];
}
}
int main()
{
scanf("%d", &t);
while (t--)
{
for (int i = 0; i < N<<1; i++) G[i].clear();
memset(dp, -1, sizeof(dp));
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
pos[a[i]] = i;
}
for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
for (int i = 1; i <= n; i++)
for (int j = max(1, b[i]-k); j <= min(n, b[i]+k); j++)
{
G[i-pos[j]+N].push_back(pos[j]);
}
for (int i = 0; i < N<<1; i++) sort(G[i].begin(), G[i].end());
int ans = f(n, n);
printf("%d\n", ans);
}
}