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 <iostream>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<iomanip>
#include<ctime>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<bitset>
#define sqr(x) ((x)*(x))
#define fz1(i,n) for ((i)=1;(i)<=(n);(i)++)
#define fd1(i,n) for ((i)=(n);(i)>=1;(i)--)
#define fz0g(i,n) for ((i)=0;(i)<=(n);(i)++)
#define fd0g(i,n) for ((i)=(n);(i)>=0;(i)--)
#define fz0k(i,n) for ((i)=0;(i)<(n);(i)++)
#define fd0k(i,n) for ((i)=(long long)((n)-1);(i)>=0;(i)--)
#define fz(i,x,y) for ((i)=(x);(i)<=(y);(i)++)
#define fd(i,y,x) for ((i)=(y);(i)>=(x);(i)--)
#define fzin fz1(i,n)
#define fzim fz1(i,m)
#define fzjn fz1(j,n)
#define fzjm fz1(j,m)
#define ff(c,itr) for (__typeof((c).begin()) itr=(c).begin();itr!=(c).end();++itr)
#define pb push_back
#define mk make_pair
#define rdst(st,len){static char ss[len];scanf("%s",ss);(st)=ss;}
#define spln(i,n) (i==n?'\n':' ')
#define fac_init(n){fac[0]=fac[1]=inv[1]=fi[0]=fi[1]=1;fz(i,2,n){fac[i]=1ll*fac[i-1]*i%mod;inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;fi[i]=1ll*fi[i-1]*inv[i]%mod;}}
using namespace std;
inline void read(int &x)
{
	char c;int f=1;
	while(!isdigit(c=getchar()))if(c=='-')f=-1;
	x=(c&15);while(isdigit(c=getchar()))x=(x<<1)+(x<<3)+(c&15);
	x*=f;
}
int t,n,m,i,j,k,a[1005][1005],ans[1005][1005],b[1005],c[1005],p[3],d[3],s[3],q[3];
string st;
void inc(int x)
{
	d[x]++;
}
void dec(int x)
{
	d[x]--;
}
void swp(int x,int y)
{
	swap(d[x],d[y]);
	int i,j;
	fz0k(i,3)if(p[i]==x)break;
	fz0k(j,3)if(p[j]==y)break;
	swap(p[i],p[j]);
}
void print()
{
	int i,j;
//	q[p[0]]=0;q[p[1]]=1;q[p[2]]=2;
	fz1(i,n)fz1(j,n){
		s[p[0]]=i;s[p[1]]=j;s[p[2]]=a[i][j];
		fz0k(k,3){
			s[k]+=d[k];
			s[k]%=n;
			while(s[k]<=0)s[k]+=n;
		}
		ans[s[0]][s[1]]=s[2];
	}
	fz1(i,n)fz1(j,n)printf("%d%c",ans[i][j],spln(j,n));
}
int main()
{
	read(t);
	while(t--)
	{
		read(n);read(m);fz1(i,n)fz1(j,n)read(a[i][j]);
		rdst(st,100005)
		fz0k(i,3){
			p[i]=i;
			d[i]=0;
		}
		fz0k(i,m){
			if(st[i]=='U'){
				dec(0);
			}
			if(st[i]=='D'){
				inc(0);
			}
			if(st[i]=='L'){
				dec(1);
			}
			if(st[i]=='R'){
				inc(1);
			}
			if(st[i]=='I'){
				swp(1,2);
			}
			if(st[i]=='C'){
				swp(0,2);
			}
//			print();
//			cerr<<endl;
		}
		print(); 
	}
	return 0;
}







