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>
#include"map"
#include"string"
#include"cmath"
#include"cstdio"
#include"iostream"
#include"cstring"
#include"algorithm"
#include"vector"
//#pragma GCC optimize(3, "Ofast")
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> pii;
inline int read() {
	int s = 0, w = 1;
	char ch = getchar();
	while (ch < '0' || ch>'9') { if (ch == '-')w = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
	return s * w;
}
#ifdef local
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif
#define f(i, l, r) for (int i = l; i <= r; ++i)
#define rf(i, r, l) for (int i = r; i >= l; --i)
#define all(x) (x).begin(), (x).end()
#define l l
#define r r
#define lch (o << 1)
#define rch (o << 1 | 1)
#define mid ((l + r) >> 1)
#define mem(x, y) memset(x, y, sizeof(x))
#define mod1(x) ((x >= mod) && (x -= mod))
#define mod2(x) ((x < 0) && (x += mod))

const int inf = 2e9 + 7;
//const ll INF = INT64_MAX;
double eps = 1e-6;
const int mod = 998244353;
const int N = 1e3 + 10;
const double pi = acos(-1.0);
ll power(ll a, ll b) { ll res = 1; while (b) { if (b & 1)res = res * a % mod; a = a * a % mod; b >>= 1; }return res; }
ll gcd(ll a, ll b) { return !b ? a : gcd(b, a % b); }
ll exgcd(ll a, ll b, ll& x, ll& y) { ll c; return !b ? (x = 1, y = 0, a) : (c = exgcd(b, a % b, y, x), y -= (a / b) * x, c); }
ll INV(ll a) { ll x, y; exgcd(a, mod, x, y); x %= mod; if (x < 0)x += mod; return x; }
/*int len, C[N], clen;
int id(int v) { return lower_bound(C + 1, C + len, v) - C; }
void initC() { sort(C + 1, C + clen+1); len = unique(C + 1, C + 1 + clen) - C - 1; }*/

int a[N][N], res[N][N];
int main() {
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#ifdef local
	freopen("in.txt", "r", stdin);
	int start = clock();
#endif
	int _; cin >> _;
	while (_--) {
		int n,m; cin >> n>>m;
		f(i, 1, n)f(j, 1, n)cin >> a[i][j];
		string s; cin >> s;
		vector<int> A(3); iota(all(A), 0);
		vector<int> B(3);
		for (auto it : s) {
			if (it == 'R') {
				B[1]++;
			}
			else if (it == 'L') {
				B[1]--;
			}
			else if (it == 'D') {
				B[0]++;
			}
			else if (it == 'U') {
				B[0]--;
			}
			else if (it == 'I') {
				swap(B[1], B[2]); swap(A[1], A[2]);
			}
			else {
				swap(B[0], B[2]); swap(A[0], A[2]);
			}
		}
		f(i, 1, n)f(j, 1, n) {
			int x = i-1, y = j-1, z = a[i][j]-1;
			vector<int> tmp,tmp2(3);
			tmp.push_back(x); tmp.push_back(y); tmp.push_back(z);
			f(k, 0, 2)tmp2[k] = (tmp[A[k]]+B[k])%n;
			for (auto& i : tmp2)if (i < 0)i += n;
			for (auto& i : tmp2)i++;
			res[tmp2[0]][tmp2[1]] = tmp2[2];
		}
		f(i, 1, n)f(j, 1, n)cout << res[i][j] << (j == n ? "\n" : " ");
	}
#ifdef local
	int end = clock();
	//cout << '\n' << end - start << "ms";
#endif* /
}