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.BufferedInputStream;
import java.util.*;

import static java.lang.Math.*;

public class C222B {

	boolean allowedDebug = true;

	public void solve() throws Exception {
		int n = ni();
		int m = ni();
		int k = ni();
		int[][] a = new int[n][m];
		for (int i = 0; i < a.length; i++) {
			for (int j = 0; j < a[0].length; j++) {
				a[i][j] = ni();
			}
		}
		List<Integer> rows = new ArrayList<Integer>();
		for (int i = 0; i < n; i++) {
			rows.add(i);
		}
		List<Integer> cols = new ArrayList<Integer>();
		for (int i = 0; i < m; i++) {
			cols.add(i);
		}
		while (k-- > 0) {
			String s = nextWord();
			int x = ni() - 1;
			int y = ni() - 1;
			switch (s.charAt(0)) {
			case 'c':
				Collections.swap(cols, x, y);
				break;
			case 'r':
				Collections.swap(rows, x, y);
				break;
			case 'g':
				println(a[rows.get(x)][cols.get(y)]);
				break;

			default:
				break;
			}
		}

	}

	// ------------------------------------------------------

	void debug(Object... o) {
		if (allowedDebug) {
			System.err.println(Arrays.deepToString(o));
		}
	}

	void print(Object... os) {
		if (os != null && os.length > 0)
			System.out.print(os[0].toString());
		for (int i = 1; i < os.length; ++i)
			System.out.print(" " + os[i].toString());
	}

	void println(Object... os) {
		print(os);
		System.out.println();
	}

	BufferedInputStream bis = new BufferedInputStream(System.in);

	String nextWord() throws Exception {
		char c = (char) bis.read();
		while (c <= ' ')
			c = (char) bis.read();
		StringBuilder sb = new StringBuilder();
		while (c > ' ') {
			sb.append(c);
			c = (char) bis.read();
		}
		return new String(sb);
	}

	String nextLine() throws Exception {
		char c = (char) bis.read();
		while (c <= ' ')
			c = (char) bis.read();
		StringBuilder sb = new StringBuilder();
		while (c != '\n' && c != '\r') {
			sb.append(c);
			c = (char) bis.read();
		}
		return new String(sb);
	}

	int ni() throws Exception {
		return Integer.parseInt(nextWord());
	}

	long nl() throws Exception {
		return Long.parseLong(nextWord());
	}

	public static void main(String[] args) throws Exception {
		new C222B().solve();
	}
}
