You are given a square matrix of size n. Every row and every column of this matrix is a permutation of 1, 2, …, n. Let a_{i, j} be the element at the intersection of i-th row and j-th column for every 1 ≤ i, j ≤ n. Rows are numbered 1, …, n top to bottom, and columns are numbered 1, …, n left to right.

There are six types of operations: 

  * R: cyclically shift all columns to the right, formally, set the value of each a_{i, j} to a_{i, ((j - 2)mod n) + 1}; 
  * L: cyclically shift all columns to the left, formally, set the value of each a_{i, j} to a_{i, (jmod n) + 1}; 
  * D: cyclically shift all rows down, formally, set the value of each a_{i, j} to a_{((i - 2)mod n) + 1, j}; 
  * U: cyclically shift all rows up, formally, set the value of each a_{i, j} to a_{(imod n) + 1, j}; 
  * I: replace the permutation read left to right in each row with its inverse. 
  * C: replace the permutation read top to bottom in each column with its inverse. 

Inverse of a permutation p_1, p_2, …, p_n is a permutation q_1, q_2, …, q_n, such that p_{q_i} = i for every 1 ≤ i ≤ n.

One can see that after any sequence of operations every row and every column of the matrix will still be a permutation of 1, 2, …, n.

Given the initial matrix description, you should process m operations and output the final matrix.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — number of test cases. t test case descriptions follow.

The first line of each test case description contains two integers n and m (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10^5) — size of the matrix and number of operations.

Each of the next n lines contains n integers separated by single spaces — description of the matrix a (1 ≤ a_{i, j} ≤ n).

The last line of the description contains a string of m characters describing the operations in order, according to the format above.

The sum of n does not exceed 1000, and the sum of m does not exceed 10^5.

Output

For each test case, print n lines with n integers each — the final matrix after m operations.

Example

Input


5
3 2
1 2 3
2 3 1
3 1 2
DR
3 2
1 2 3
2 3 1
3 1 2
LU
3 1
1 2 3
2 3 1
3 1 2
I
3 1
1 2 3
2 3 1
3 1 2
C
3 16
1 2 3
2 3 1
3 1 2
LDICRUCILDICRUCI


Output


2 3 1 
3 1 2 
1 2 3 

3 1 2 
1 2 3 
2 3 1 

1 2 3 
3 1 2 
2 3 1 

1 3 2 
2 1 3 
3 2 1 

2 3 1 
3 1 2 
1 2 3

Note

Line breaks between sample test case answers are only for clarity, and don't have to be printed.
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<queue>
#define fs(i,x,y) for(int i=(x),_=(y);i<=_;++i)
#define fn(i,x,y) for(int i=(x),_=(y);i>=_;--i)
#define tor(i,v,x) for(int i=head[x],v=to[i];i;i=nxt[i],v=to[i])
#define ls(x) ch[x][0]
#define rs(x) ch[x][1]
#define mpi(x,y) make_pair(x,y)
#define pi pair<int,int>
#define DEBUG printf("%s %d\n",__FUNCTION__,__LINE__)
using namespace std;
typedef long long ll;
typedef double db;
template<typename T>
void read(T &x){
	x=0;char ch=getchar();bool f=0;
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x=f?-x:x;
}

const int N=1e3+7,M=1e5+7;
int n,m;
int a[N][N];
int b[N][N];
char s[M];

int chk(int tag,int i,int j){
	if(tag==1)return i;
	else if(tag==2)return j;
	else return a[i][j];
}

signed main(){
	int T;read(T);while(T--){
		read(n);read(m);fs(i,0,n-1)fs(j,0,n-1)read(a[i][j]),--a[i][j];
		scanf("%s",s+1);
		//神仙题，反演操作相当于坐标反转
		int t1=1,t2=2,t3=3;
		int d1=0,d2=0,d3=0;
		fs(i,1,m){
			if(s[i]=='R')++d2;
			else if(s[i]=='L')--d2;
			else if(s[i]=='U')--d1;
			else if(s[i]=='D')++d1;
			else if(s[i]=='I')swap(d2,d3),swap(t2,t3);
			else if(s[i]=='C')swap(d1,d3),swap(t1,t3);
		}
		d1=(d1%n+n)%n;d2=(d2%n+n)%n;d3=(d3%n+n)%n;
		//t1t2t3表示是原来哪一维
		fs(i,0,n-1)fs(j,0,n-1){
			int x=(chk(t1,i,j)+d1)%n;
			int y=(chk(t2,i,j)+d2)%n;
			int z=(chk(t3,i,j)+d3)%n;
			b[x][y]=z+1;
		}
		fs(i,0,n-1)fs(j,0,n-1)printf("%d%c",b[i][j]," \n"[j==n-1]);
	}
	return 0;
}