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 static java.util.Arrays.deepToString;

import java.io.*;
import java.util.*;

public class B {

    static void solve() throws Exception {
        int n = nextInt();
        int m = nextInt();
        int k = nextInt();
        int[][] a = new int[n + 1][m + 1];
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                a[i][j] = nextInt();
            }
        }
        int[] mx = new int[1001];
        for(int i = 1; i <= n; i++){
            a[i][0] = i;
            mx[i] = i;
        }
        int[] my = new int[1001];
        for(int i = 1; i <= m; i++){
            a[0][i] = i;
            my[i] = i; 
        }
//      1 1 m[1] = 1
//      2 2
//      3 3 m[3] = 3
//      
//      3 1 m[3] = 1
//      2 2
//      1 3 m[1] = 3
        for(int i = 0; i < k; i++){
            char cmd = next().charAt(0);
            int x = nextInt();
            int y = nextInt();
            if(cmd == 'c'){
                int tmp = a[0][x];
                a[0][x] = a[0][y];
                a[0][y] = tmp;
                tmp = my[x];
                my[x] = my[y];
                my[y] = tmp;
            }else if(cmd == 'r'){
                int tmp = a[x][0];
                a[x][0] = a[y][0];
                a[y][0] = tmp;
                tmp = mx[x];
                mx[x] = mx[y];
                mx[y] = tmp;
            }else{
                int c1 = mx[x];
                int c2 = my[y];
                System.out.println(a[c1][c2]);
            }
        }
    }

    static BufferedReader br;
    static StringTokenizer st;
    static PrintWriter out;

    static void debug(Object... a) {
        System.err.println(deepToString(a));
    }

    static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            String line = br.readLine();
            if (line == null) {
                return null;
            }
            st = new StringTokenizer(line);
        }
        return st.nextToken();
    }

    static long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    static double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    public static void main(String[] args) {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            solve();
            br.close();
        } catch (Throwable e) {
            e.printStackTrace();
            System.exit(1);
        }
    }
}