The Free Meteor Association (FMA) has got a problem: as meteors are moving, the Universal Cosmic Descriptive Humorous Program (UCDHP) needs to add a special module that would analyze this movement. 

UCDHP stores some secret information about meteors as an n × m table with integers in its cells. The order of meteors in the Universe is changing. That's why the main UCDHP module receives the following queries:

  * The query to swap two table rows; 
  * The query to swap two table columns; 
  * The query to obtain a secret number in a particular table cell. 



As the main UCDHP module is critical, writing the functional of working with the table has been commissioned to you.

Input

The first line contains three space-separated integers n, m and k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 500000) — the number of table columns and rows and the number of queries, correspondingly.

Next n lines contain m space-separated numbers each — the initial state of the table. Each number p in the table is an integer and satisfies the inequality 0 ≤ p ≤ 106.

Next k lines contain queries in the format "si xi yi", where si is one of the characters "с", "r" or "g", and xi, yi are two integers.

  * If si = "c", then the current query is the query to swap columns with indexes xi and yi (1 ≤ x, y ≤ m, x ≠ y); 
  * If si = "r", then the current query is the query to swap rows with indexes xi and yi (1 ≤ x, y ≤ n, x ≠ y); 
  * If si = "g", then the current query is the query to obtain the number that located in the xi-th row and in the yi-th column (1 ≤ x ≤ n, 1 ≤ y ≤ m). 



The table rows are considered to be indexed from top to bottom from 1 to n, and the table columns — from left to right from 1 to m.

Output

For each query to obtain a number (si = "g") print the required number. Print the answers to the queries in the order of the queries in the input.

Examples

Input

3 3 5
1 2 3
4 5 6
7 8 9
g 3 2
r 3 2
c 2 3
g 2 2
g 3 2


Output

8
9
6


Input

2 3 3
1 2 4
3 1 5
c 2 1
r 1 2
g 1 3


Output

5

Note

Let's see how the table changes in the second test case.

After the first operation is fulfilled, the table looks like that:

2 1 4

1 3 5

After the second operation is fulfilled, the table looks like that:

1 3 5

2 1 4

So the answer to the third query (the number located in the first row and in the third column) will be 5.
import java.io.*;
import java.util.*;

public class B {
	BufferedReader in;
	PrintWriter out;
	StringTokenizer st;

	void solve() throws IOException {
		int n, m , k;
		n = nextInt();
		m = nextInt();
		k = nextInt();
		int[][] a = new int[n][m];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				a[i][j] = nextInt();
		int[] clm = new int[m];
		int[] str = new int[n];
		for (int i = 0; i < m; i++)
			clm[i] = i;
		for (int i = 0; i < n; i++)
			str[i] = i;
		String s ="";
		for (int i = 0; i < k; i++){
			s = nextToken();
			int x, y;
			x = nextInt() - 1;
			y = nextInt() - 1;
			//swap clms
			if (s.equals("c")){	
				swap(clm, x, y);
			}
			if (s.equals("r")){	
				swap(str, x, y);
			}
			if (s.equals("g")){	
				out.println(a[str[x]][clm[y]]);
			}
		}
	}

	void swap(int[] a, int x, int y){
		int c = a[x];
		a[x] = a[y];
		a[y] = c;
	}
	
	void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(new OutputStreamWriter(System.out));

		solve();

		in.close();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new B().run();
	}

	String nextToken() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(in.readLine());
		}
		return st.nextToken();
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

}