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 <bits/stdc++.h>
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define V vector
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define ALL(v) (v).begin(), (v).end()
#define debug(x) cerr << "Line(" << __LINE__ << ") -> " << #x << " is " << x << endl

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;

const int INF = 1e9 + 7;

int ans[1003][1003];

signed main()
{
	IO_OP;

	int t;
	cin >> t;
	while(t--) {
		int n, m;
		cin >> n >> m;
		V<array<int, 3>> v;
		array<int, 3> a = {0, 1, 2}, b = {0, 0, 0};
		for(int i = 0; i < n; i++)
			for(int j = 0; j < n; j++) {
				int x;
				cin >> x;
				x--;
				v.PB({i, j, x});
			}
		string s;
		cin >> s;
		for(char c:s) {
			if(c == 'R') b[1] = (b[1] + 1) % n;
			if(c == 'L') b[1] = (b[1] + n - 1) % n;
			if(c == 'D') b[0] = (b[0] + 1) % n;
			if(c == 'U') b[0] = (b[0] + n - 1) % n;
			if(c == 'I') swap(a[1], a[2]), swap(b[1], b[2]);
			if(c == 'C') swap(a[0], a[2]), swap(b[0], b[2]);
		}
		for(auto p:v)
			ans[(p[a[0]] + b[0]) % n][(p[a[1]] + b[1]) % n] = (p[a[2]] + b[2]) % n + 1;
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < n; j++)
				cout << ans[i][j] << " ";
			cout << endl;
		}
	}
	
}

