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

public class Main {
    
    public static void main(String[] args) throws IOException {
        PrintWriter out = new PrintWriter(System.out);
        FastReader in = new FastReader(new BufferedReader(new InputStreamReader(System.in)));
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }
    
}

class Task {
     
    public void solve(FastReader in, PrintWriter out) throws IOException {
        int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
        int pi[] = new int[n], pj[] = new int[m];
        int a[][] = new int[n][m];
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < m; ++j) {
                a[i][j] = in.nextInt();
            }
        }
        for(int i = 0; i < n; ++i) {
            pi[i] = i;
        }
        for(int j = 0; j < m; ++j) {
            pj[j] = j;
        }
        for(int i = 0; i < k; ++i) {
            String s = in.next();
            int x = in.nextInt() - 1, y = in.nextInt() - 1;
            if(s.charAt(0) == 'c') {
                int tmp = pj[x];
                pj[x] = pj[y];
                pj[y] = tmp;
            } else if(s.charAt(0) == 'r') {
                int tmp = pi[x];
                pi[x] = pi[y];
                pi[y] = tmp;
            } else {
                out.println(a[pi[x]][pj[y]]);
            }
        }
    }
    
}

class FastReader {
    
    BufferedReader in;
    StringBuilder sb = new StringBuilder();
    
    public FastReader(BufferedReader in) {
        this.in = in;
    }
    
    public String next() throws IOException {
        sb.setLength(0);
        while(true) {
            int c = in.read();
            if(c == -1) {
                return null;
            }
            if(" \n\r\t".indexOf(c) == -1) {
                sb.append((char)c);
                break;
            }
        }
        while(true) {
            int c = in.read();
            if(c == -1 || " \n\r\t".indexOf(c) != -1) {
                break;
            }
            sb.append((char)c);
        }
        return sb.toString();
    }
    
    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    
    public long nextLong() throws IOException {
        return Long.parseLong(next());
    }
    
    public double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
    
}
