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.awt.*;
import java.awt.event.*;
import java.awt.geom.*;
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;

import static java.lang.Double.parseDouble;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.util.Arrays.binarySearch;
import static java.util.Arrays.sort;
import static java.util.Collections.binarySearch;
import static java.util.Collections.shuffle;
import static java.util.Collections.sort;

public class B {
	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;

	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(
				System.out)));
		int r = readInt();
		int c = readInt();
		int qq = readInt();
		int[][] grid = new int[r][c];
		for(int i = 0; i < r; i++)	{
			for(int j = 0; j < c; j++)	{
				grid[i][j] = readInt();
			}
		}
		int[] row = new int[r];
		int[] col = new int[c];
		for(int i = 0; i < r; i++)	{
			row[i] = i;
		}
		for(int i = 0; i < c; i++)	{
			col[i] = i;
		}
		while(qq-- > 0)	{
			String type = nextToken();
			int x = readInt();
			int y = readInt();
			x--;
			y--;
			if(type.equals("r"))	{
				int temp = row[x];
				row[x] = row[y];
				row[y] = temp;
			}
			else if(type.equals("c"))	{
				int temp = col[x];
				col[x] = col[y];
				col[y] = temp;
			}
			else	{
				pw.println(grid[row[x]][col[y]]);
			}
		}
		pw.close();
	}

	public static long readLong() throws IOException {
		return parseLong(nextToken());
	}

	public static double readDouble() throws IOException {
		return parseDouble(nextToken());
	}

	public static int readInt() throws IOException {
		return parseInt(nextToken());
	}

	public static String nextToken() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			if (!br.ready()) {
				pw.close();
				System.exit(0);
			}
			st = new StringTokenizer(br.readLine());
		}
		return st.nextToken();
	}
}
