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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.*;
public class Main {

	/**
	 * @param args
	 */
	private static void swap(int[] a, int i, int j)
	{
		int tmp = a[i];
		a[i] = a[j];
		a[j] = tmp;
	}
	public static void main(String[] args) throws IOException{
		// TODO Auto-generated method stub
		Reader kb = new Reader(System.in);
		int r = kb.nextInt();
		int c= kb.nextInt();
		int a[][] =new int[r][c];
		int n = kb.nextInt();
		for(int i =0;i<r;i++)
			for(int j =0;j<c;j++)
				a[i][j]=kb.nextInt();
		int rr[] = new int[r];
		for(int i =0;i<r;i++)
			rr[i]=i;
		int cc[] = new int[c];
		for(int i =0;i<c;i++)
			cc[i]=i;
		for(int k =0;k<n;k++){
			String temp = kb.nextString();
			if(temp.equals("r"))
				swap(rr,kb.nextInt()-1,kb.nextInt()-1);
			else if(temp.equals("c"))
				swap(cc,kb.nextInt()-1,kb.nextInt()-1);
			else if(temp.equals("g"))
				System.out.println(a[rr[kb.nextInt()-1]][cc[kb.nextInt()-1]]);
		}
	}

}
class Reader
{
	private BufferedReader x;
	private StringTokenizer st;
	
	public Reader(InputStream in)
	{
		x = new BufferedReader(new InputStreamReader(in));
		st = null;
	}
	public String nextString() throws IOException
	{
		while( st==null || !st.hasMoreTokens() )
			st = new StringTokenizer(x.readLine());
		return st.nextToken();
	}
	public int nextInt() throws IOException
	{
		return Integer.parseInt(nextString());
	}
	public double nextDouble() throws IOException
	{
		return Double.parseDouble(nextString());
	}
}
