You are given an array a of length n.

You are also given a set of distinct positions p1,p2,…,pm, where 1≤pi<n. The position pi means that you can swap elements a[pi] and a[pi+1]. You can apply this operation any number of times for each of the given positions.

Your task is to determine if it is possible to sort the initial array in non-decreasing order (a1≤a2≤?≤an) using only allowed swaps.

For example, if a=[3,2,1] and p=[1,2], then we can first swap elements a[2] and a[3] (because position 2 is contained in the given set p). We get the array a=[3,1,2]. Then we swap a[1] and a[2] (position 1 is also contained in p). We get the array a=[1,3,2]. Finally, we swap a[2] and a[3] again and get the array a=[1,2,3], sorted in non-decreasing order.

You can see that if a=[4,1,2,3] and p=[3,2] then you cannot sort the array.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤100) — the number of test cases.

Then t test cases follow. The first line of each test case contains two integers n and m (1≤m<n≤100) — the number of elements in a and the number of elements in p. The second line of the test case contains n integers a1,a2,…,an (1≤ai≤100). The third line of the test case contains m integers p1,p2,…,pm (1≤pi<n, all pi are distinct) — the set of positions described in the problem statement.

Output
For each test case, print the answer — “YES” (without quotes) if you can sort the initial array in non-decreasing order (a1≤a2≤?≤an) using only allowed swaps. Otherwise, print “NO”.

Example
Input
6
3 2
3 2 1
1 2
4 2
4 1 2 3
3 2
5 1
1 2 3 4 5
1
4 2
2 1 4 3
1 3
4 2
4 3 2 1
1 3
5 2
2 1 2 3 3
1 4
Output
YES
NO
YES
YES
NO
YES
總感覺這個(gè)題有什么比較好的解法的，但是我太菜了，沒想到。我的做法就是直接暴力。因?yàn)閿?shù)據(jù)量不大，感覺循環(huán)幾次就可以判斷出能否實(shí)現(xiàn)。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=1e2+100; int a[maxx]; int n,m;inline bool check() {int flag=1;for(int i=1;i<n;i++){if(a[i]>a[i+1]) return 0;}return 1; } int main() {int t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++) scanf("%d",&a[i]);set<int> s;int x;for(int i=1;i<=m;i++) {scanf("%d",&x);s.insert(x);}int flag=1;while(1){if(check()) break;for(int i=1;i<n;i++){if(a[i]>a[i+1]){set<int> ::iterator it=s.lower_bound(i);if(*it!=i){flag=0;break;}swap(a[i],a[i+1]);}}if(flag==0) break;//for(int i=1;i<=n;i++) cout<<a[i]<<" ";cout<<endl;}if(flag) cout<<"YES"<<endl;else cout<<"NO"<<endl;}return 0; }

努力加油a啊，(^o)/~

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