一，题目

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a permutation p of length n, an array of m distinct integers a1,a2,…,am (1≤ai≤n), and an integer d.

Let pos(x) be the index of x in the permutation p. The array a is not good if

• pos(ai)<pos(ai+1)≤pos(ai)+d for all 1≤i<m.

For example, with the permutation p=[4,2,1,3,6,5] and d=2:

• a=[2,3,6] is a not good array.

• a=[2,6,5] is good because pos(a1)=2, pos(a2)=5, so the condition pos(a2)≤pos(a1)+d is not satisfied.

• a=[1,6,3] is good because pos(a2)=5, pos(a3)=4, so the condition pos(a2)<pos(a3) is not satisfied.

In one move, you can swap two adjacent elements of the permutation p. What is the minimum number of moves needed such that the array a becomes good? It can be shown that there always exists a sequence of moves so that the array a becomes good.

A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3, but there is 4 in the array).

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤104). The description of the test cases follows.

The first line of each test case contains three integers n, m and d (2≤n≤10^5, 2≤m≤n, 1≤d≤n), the length of the permutation p, the length of the array a and the value of d.

The second line contains n integers p1,p2,…,pn (1≤pi≤n, pi≠pj for i≠j).

The third line contains m distinct integers a1,a2,…,am (1≤ai≤n, ai≠aj for i≠j).

The sum of n over all test cases doesn't exceed 5*10^5.

Output

For each test case, print the minimum number of moves needed such that the array a becomes good.

Example

Input

5

4 2 2

1 2 3 4

1 3

5 2 4

5 4 3 2 1

5 2

5 3 3

3 4 1 5 2

3 1 2

2 2 1

1 2

2 1

6 2 4

1 2 3 4 5 6

2 5

Output

1

3

2

0

2

Note

In the first case, pos(a1)=1, pos(a2)=3. To make the array good, one way is to swap p3 and p4. After that, the array a will be good because the condition pos(a2)≤pos(a1)+d won't be satisfied.

In the second case, pos(a1)=1, pos(a2)=4. The 3 moves could be:

1. Swap p3 and p4.

1. Swap p2 and p3.

1. Swap p1 and p2.

After these moves, the permutation p will be [2,5,4,3,1]. The array a will be good because the condition pos(a1)<pos(a2) won't be satisfied. It can be shown that you can't make the array a good with fewer moves.

In the third case, pos(a1)=1, pos(a2)=3, pos(a3)=5. The 2 moves can be:

1. Swap p4 and p5.

1. Swap p3 and p4.

After these moves, the permutation p will be [3,4,2,1,5]. The array a will be good because the condition pos(a2)<pos(a3) won't be satisfied. It can be shown that you can't make the array a good with fewer moves.

In the fourth case, pos(a1)=2, pos(a2)=1. The array a is already good.

In the fifth case, pos(a1)=2, pos(a2)=5. The 2 moves are:

1. Swap p1 and p2.

1. Swap p5 and p6.

二，思路

（1）因为只要保证a数组中有一对和不满足pos()<pos()≤pos()+d就能够满足答案要求，所以最优方案肯定只改变一对和在p数组中的相对位置（pos()即ai在p数组中的位置），或者无须改变任何操作（可能a数组已经有一对和不满足pos()<pos()≤pos()+d，所以接下来的思路就是如何改变和在p数组中的相对位置使得交换数组p中数字的次数最小。

（2）首先，思路一，让pos()>pos()，仔细一想，就是让在p中的相对位置在右边，说白了就是一直与其他数交换，直到换的的左边。这时，想要交换次数最小，我们就需要定义一个变量MIN，用来找到所有和的中在p数组中的最小相对距离。根据这个思路，我们可以得到最小的交换次数为MIN。

（3）其次，思路二，让pos()>pos()+d,当然，这个是不可能的，pos()pos()+d,即pos()-pos()d,说白了，就是的相对位置在左边，且距离至少为d。这时，想要交换次数最小，我们就需要定义一个变量MAX，用来找到所有和的中在p数组中的最大相对距离。根据这个思路，我们可以得到的最小交换次数为d-MAX+1(不知道为啥要加1，反正当时看到自己运行结果和答案少了个一就加了一)。

（4）然后，就是一些特判，首先，如果MIN为负数的话，就说明肯定存在一对和，使得pos()-pos(）>0,既然这样，那就不用交换了，即交换次数为0。其次，如果MAX>d，就说明pos()-pos(）>=d,也是不用交换。最后，就是如果d>n-1的话，pos()-pos(）将不可能大于等于d，所以这种情况下只能用思路一。

（5）那么什么情况下用思路一或二呢，什么情况都用，到最后得到MIN和MAX，我们就判断一下思路一的最小交换次数和思路二哪个交换次数小，然后输出小的。当然，d>n-1的话只能输出MIN。

三，代码

#include<iostream>
#include<cmath>
using namespace std;
#define int long long
#define endl '\n'
int p[200005]={0},a[200005]={0},book[200005]={0};
signed main()
{ios::sync_with_stdio(false); cin.tie(0),cout.tie(0);int t;cin>>t;while(t--){int n,m,d;cin>>n>>m>>d;int MIN=n+1,MAX=-n-1;for(int i=0;i<n;++i) cin>>p[i],book[p[i]]=i+1;//book用于记录a[i]在数组p中的相对距离for(int i=0;i<m;++i) {cin>>a[i];if(i!=0){MIN=min(book[a[i]]-book[a[i-1]],MIN);MAX=max(book[a[i]]-book[a[i-1]],MAX);}}if(MAX>d||MIN<0) cout<<0<<endl;else{if(d-MAX+1<=MIN&&d<n-1) cout<<d-MAX+1<<endl;else  cout<<MIN<<endl;}}return 0;
}