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.geom.*;
import java.io.*;
import java.math.*;
import java.util.*;
import java.util.regex.*;
import static java.lang.Math.*;

public class Main {
	public Main() throws IOException {
		String input;

		input = br.readLine();
		work(input, 0);

		System.out.print(sb);
	}

	private void work(String input, int t) throws IOException {
		//debug(t, input);
		String[] inputArray = input.split(" ");
		int n = Integer.valueOf(inputArray[0]);
		int m = Integer.valueOf(inputArray[1]);
		int k = Integer.valueOf(inputArray[2]);
		int[][] map = new int[n][m];
		int[] mapR = new int[n];
		int[] mapC = new int[m];

		for (int i = 0; i < n; i++) {
			input = br.readLine();
			inputArray = input.split(" ");
			for (int j = 0; j < m; j++) {
				map[i][j] = Integer.valueOf(inputArray[j]);
			}
		}

		for (int i = 0; i < n; i++)
			mapR[i] = i;

		for (int i = 0; i < m; i++)
			mapC[i] = i;

		for (int i = 0; i < k; i++) {
			input = br.readLine();
			inputArray = input.split(" ");
			int index1 = Integer.valueOf(inputArray[1]) - 1;
			int index2 = Integer.valueOf(inputArray[2]) - 1;
			int temp;

			if (inputArray[0].charAt(0) == 'c') {
				temp = mapC[index1];
				mapC[index1] = mapC[index2];
				mapC[index2] = temp;
			} else if (inputArray[0].charAt(0) == 'r') {
				temp = mapR[index1];
				mapR[index1] = mapR[index2];
				mapR[index2] = temp;
			} else {
				sb.append(map[mapR[index1]][mapC[index2]]);
				sb.append("\n");
			}
		}
	}

	StringBuilder sb = new StringBuilder();
	BufferedReader br = new BufferedReader(
		new InputStreamReader(System.in));

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

	public static void debug(Object... arr) {
		System.err.println(Arrays.deepToString(arr));
	}
}
