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.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class Solver {
	
	StringTokenizer Stok;
	PrintWriter pw;
	BufferedReader br;
	
	public static void main(String[] Args) throws NumberFormatException, IOException{
		new Solver().Solve();
	}
	
	public String nextString() throws IOException{
		while (Stok==null || !Stok.hasMoreTokens()){
			Stok=new StringTokenizer(br.readLine());
		}
		return Stok.nextToken();
	}
	
	public int nextInt() throws NumberFormatException, IOException{
		return Integer.parseInt(nextString());
	}
	
	public long nextLong() throws NumberFormatException, IOException{
		return Long.parseLong(nextString());
	}
	
	public double nextDouble() throws NumberFormatException, IOException{
		return Double.parseDouble(nextString());
	}
	
	int n,m,k;
	int[][] Mas;
	int[] Rows, Kols;
	
	public void Solve() throws NumberFormatException, IOException{
		//pw=new PrintWriter("output.txt");
		pw=new PrintWriter(new OutputStreamWriter(System.out));
		//br=new BufferedReader(new FileReader("input.txt"));
		br=new BufferedReader(new InputStreamReader(System.in));
		n=nextInt();
		m=nextInt();
		k=nextInt();
		
		Rows=new int[n+1];
		Kols=new int[m+1];
		
		for (int i=0; i<n; i++)
			Rows[i]=i+1;
		
		for (int i=0; i<m; i++)
			Kols[i]=i+1;
		
		Mas=new int[n+1][m+1];
		for (int i=0; i<n; i++){
			for (int j=0; j<m; j++){
				Mas[i][j]=nextInt();
			}
		}
		
		char typecomm;
		int buf=0;
		int x,y=0;
		for (int i=0; i<k; i++){
			typecomm=nextString().charAt(0);
			x=nextInt()-1;
			y=nextInt()-1;
			switch (typecomm){
			case 'c':
				buf=Kols[x];
				Kols[x]=Kols[y];
				Kols[y]=buf;
				break;
			case 'r':
				buf=Rows[x];
				Rows[x]=Rows[y];
				Rows[y]=buf;
				; break;
			case 'g':
				pw.println(Mas[Rows[x]-1][Kols[y]-1]);
				; break;
			}
		}
		pw.flush();
		pw.close();
	}
}
