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.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.StringTokenizer;

public class Solution {
    //<editor-fold desc="input parse" defaultstate="collapsed">

    private static StringTokenizer st;
    private static BufferedReader reader;
    private static BufferedWriter writer;

    private static long nextLong() {
        return Long.parseLong(st.nextToken());
    }

    private static int nextInt() {
        return Integer.parseInt(st.nextToken());
    }

    private static double nextDouble() {
        return Double.parseDouble(st.nextToken());
    }

    private static short nextShort() {
        return Short.parseShort(st.nextToken());
    }

    private static byte nextByte() {
        return Byte.parseByte(st.nextToken());
    }

    private static void initTokenizer() throws Exception {
        st = new StringTokenizer(reader.readLine());
    }

    //</editor-fold>
    

    public static void main(String[] args) throws Exception {
	reader = new BufferedReader(new InputStreamReader(System.in));
	writer = new BufferedWriter(new OutputStreamWriter(System.out), 1 << 20);
	
	initTokenizer();
	
	int n = nextInt();
	int m = nextInt();
	
	int queries = nextInt();
	
	int[] rows = new int[n];
	int[] cols = new int[m];
	
	for (int i = 0; i < rows.length; i++) rows[i] = i;
	for (int i = 0; i < cols.length; i++) cols[i] = i;
	
	int[][] a = new int[n][m];
	
	for (int i = 0; i < n; i++) {
	    initTokenizer();
	    
	    for (int j = 0; j < m; j++) {
		a[i][j] = nextInt();
	    }
	}
	
	int swap;
	
	for (int q = 0; q < queries; q++) {
	    initTokenizer();
	    String que = st.nextToken();
	    
	    int x = nextInt() - 1;
	    int y = nextInt() - 1;
	    
	    switch (que) {
		case "c": {
		    swap = cols[x];
		    cols[x] = cols[y];
		    cols[y] = swap;
		    break;
		}
		case "r": {
		    swap = rows[x];
		    rows[x] = rows[y];
		    rows[y] = swap;
		    break;
		}
		case "g": {
		    writer.write(String.valueOf(a[rows[x]][cols[y]]));
		    writer.newLine();
		    break;
		}
	    }
	}
	
	writer.close();
    }
}