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.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Main
{


    public static void main(String[] args) throws IOException
    {
        FastScanner input = new FastScanner();
		
        
      Integer n = input.nextInt();
      Integer m =input.nextInt();
      Integer q =input.nextInt();
      Integer a[][] = new Integer[n+1][m+1];
        for (int i = 1; i <=n; i++) {
            
            for (int j = 1; j <=m; j++) {
                a[i][j] =  input.nextInt();
            }
        }
        StringBuilder ans = new StringBuilder();
        Integer R[] = new Integer[n+1];
        for (int i = 1; i <=n; i++) {
            R[i]=i;
        }
        Integer C[] = new Integer[m+1];
        for (int i = 1; i <=m; i++) {
            C[i]=i;
        }
        for (int i = 0; i < q; i++) {
           
            char ch = input.next().charAt(0);
            Integer one = input.nextInt();
            Integer two = input.nextInt();
            if(ch=='c')
            {
                 Integer temp = C[one];
                C[one] = C[two];
                C[two] = temp;
            }else if(ch=='r')
            {
                Integer temp = R[one];
                R[one] = R[two];
                R[two] = temp;
            }else
            {
                ans.append(a[R[one]][C[two]]+"\n");
            }
        }
        System.out.println(ans);
        
        
    }
    static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			
			return Long.parseLong(next());
		}
	}

}
