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 javax.jws.Oneway;
import java.io.*;
import java.math.BigInteger;
import static java.math.BigInteger.*;

import java.security.PublicKey;
import java.util.*;

public class Solution{



    void solve()throws Exception
    {


        int n=nextInt();
        int m=nextInt();
        int k=nextInt();
        int[][]a=new int[n][m];
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
                a[i][j]=nextInt();
        int[][]p=new int[2][];

        p[0]=new int[n];
        p[1]=new int[m];

        int[][]q=new int[2][];

        q[0]=new int[n];
        q[1]=new int[m];
        for(int i=0;i<n;i++)
            p[0][i]=q[0][i]=i;
        for(int i=0;i<m;i++)
            p[1][i]=q[1][i]=i;
        for(int step=0;step<k;step++)
        {

            String s=nextString();
            int x=nextInt()-1;
            int y=nextInt()-1;
            if(s.equals("g"))
                writer.println(a[p[0][x]][p[1][y]]);
            else if(s.equals("r"))
            {
                swap(p[0],q[0],x,y);
            }
            else
                swap(p[1],q[1],x,y);


        }

    }

    private void swap(int[] p, int[] q, int x, int y) {
        int temp=p[x];
        p[x]=p[y];
        p[y]=temp;

    }

    ////////////
    BufferedReader reader;
    PrintWriter writer;
    StringTokenizer stk;
    void run()throws Exception
    {

        reader=new BufferedReader(new InputStreamReader(System.in));
        //reader=new BufferedReader(new FileReader("3.in"));
        stk=null;
        writer=new PrintWriter(new PrintWriter(System.out));
        //writer=new PrintWriter(new FileWriter("output.txt"));
        solve();
        reader.close();
        writer.close();
    }
    int nextInt()throws Exception
    {
        return Integer.parseInt(nextToken());
    }

    long nextLong()throws Exception
    {
        return Long.parseLong(nextToken());

    }
    double nextDouble()throws Exception
    {
        return Double.parseDouble(nextToken());


    }

    String nextString()throws Exception
    {
        return nextToken();
    }
    String nextLine()throws Exception
    {
        return reader.readLine();
    }
    String nextToken()throws Exception
    {
        if(stk==null || !stk.hasMoreTokens())
        {
            stk=new StringTokenizer(nextLine());
            return nextToken();

        }
        return stk.nextToken();
    }

    public static void main(String[]args) throws Exception
    {
        new Solution().run();
    }








}
