The only difference between easy and hard versions is constraints.
There are nn kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the pi-th kid (in case of i=pi the kid will give his book to himself). It is guaranteed that all values of pipi are distinct integers from 1to nn (i.e. p is a permutation). The sequence pp doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4,6,1,3,5,2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p=[5,1,2,4,3]. The book of the 1-st kid will be passed to the following kids:
after the 1-st day it will belong to the 5-th kid,after the 2-nd day it will belong to the 3-rd kid,after the 3-rd day it will belong to the 2-nd kid,after the 4-th day it will belong to the 1-st kid.So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1≤q≤200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1≤n≤200) — the number of kids in the query. The second line of the query contains n integers p1,p2,…,pn (1≤pi≤n, all pi are distinct, i.e. p is a permutation), where pi is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a1,a2,…,an, where aiai is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6 5 1 2 3 4 5 3 2 3 1 6 4 6 2 1 5 3 1 1 4 3 4 1 2 5 5 1 2 4 3Output
1 1 1 1 1 3 3 3 2 3 3 2 1 3 1 2 2 2 2 4 4 4 1 4这一题的题意是给出q组示例，每组2行，第一行为一个数n，接着给出这n个数，
讲的是经过多少次轮换，自己的书又回到自己手上，每个人手上都有一个数，被转过来的书，就会转到自己手上那个数指的人身上。如此反复，求每个人的书转了多少次，才回到自己手上。
这个是简单版本的，所以直接用搜索就可以求出来的。
代码如下：
#include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #define ll long long using namespace std; int y=0; int a[1000010]; void dfs(int x,int z) { if(x==z) return ; y++; dfs(a[x],z); } int main() { int t; scanf("%d",&t); while(t--) { int m; scanf("%d",&m); for(int i=1;i<=m;i++) scanf("%d",&a[i]); for(int i=1;i<=m;i++) { y=1; dfs(a[i],i); printf("%d",y); if(i!=m) printf(" "); } printf("\n"); } return 0; }
