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.
//Date: Oct 12, 2013 
//Time: 4:49:19 PM

import java.util.*;
import java.io.*;

public class B222 implements Runnable {

	public void solve() throws IOException {
		int M = nextInt();
		int N = nextInt();
		int Q = nextInt();
		int[][] a = new int[M][N];
		for(int i = 0; i < M; i++)
			for(int j = 0; j < N; j++) a[i][j] = nextInt();
		
		int[] r = new int[M];
		for(int i = 0; i < M; i++) r[i] = i;
		
		int[] c = new int[N];
		for(int i = 0; i < N; i++) c[i] = i;
		StringBuilder sb = new StringBuilder();
		
		for(int i = 0;i < Q; i++){
			char task = nextToken().charAt(0);
			int rr = nextInt() - 1;
			int cc = nextInt() - 1;
			
			if(task == 'r'){
				int t = r[rr];
				r[rr] = r[cc];
				r[cc] = t;
			}
			else if(task == 'c'){
				int t = c[rr];
				c[rr] = c[cc];
				c[cc] = t;
			}
			else{
				sb.append(a[r[rr]][c[cc]] + "\n");
			}
		}
		
		System.out.print(sb.toString());
	}

	
	
	
	
	
	
	
	
	
	
	
	//-----------------------------------------------------------
	public static void main(String[] args) {
		new B222().run();
	}

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			tok = null;
			solve();
			in.close();
		} catch (IOException e) {
			System.exit(0);
		}
	}

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

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

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

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

	BufferedReader in;
	StringTokenizer tok;
}
/**
public class B222 {

}
 
*/