Ashish has n elements arranged in a line.

These elements are represented by two integers ai — the value of the element and bi — the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of ai.

He can perform the following operation any number of times:

Select any two elements i and j such that bi≠bj and swap them. That is, he can only swap two elements of different types in one move.
Tell him if he can sort the elements in non-decreasing values of ai after performing any number of operations.

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

The first line of each test case contains one integer n (1≤n≤500) — the size of the arrays.

The second line contains n integers ai (1≤ai≤105) — the value of the i-th element.

The third line containts n integers bi (bi∈{0,1}) — the type of the i-th element.

Output
For each test case, print “Yes” or “No” (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.

You may print each letter in any case (upper or lower).

Example
Input
5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1
Output
Yes
Yes
Yes
No
Yes
Note
For the first case: The elements are already in sorted order.

For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.

For the third case: The elements are already in sorted order.

For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that bi≠bj. The elements cannot be sorted.

For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2.
思路：我們可以找規律發現，只有序列中同時包含0或者1的話，就一定可以變得符合題意。如果只有一個數字的話，必須是非遞減序列才可以。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=521; struct node{int val;int mk;int id;bool operator<(const node &a)const{if(val!=a.val) return val<a.val;else return id<a.id;} }p[maxx]; int n;int main() {int t;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&p[i].val),p[i].id=i;for(int i=1;i<=n;i++) scanf("%d",&p[i].mk);if(n==1) {cout<<"YES"<<endl;continue;}int i;for(i=2;i<=n;i++) if(p[i].mk!=p[i-1].mk) break;if(i>n){int flag=1;for(int j=2;j<=n;j++){if(p[j].val<p[j-1].val){flag=0;break;}}if(flag) cout<<"YES"<<endl;else cout<<"NO"<<endl;}else cout<<"YES"<<endl;}return 0; }

努力加油a啊，(^o)/~

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