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

/* Name of the class has to be "Main" only if the class is public. */
public class Ideone
{ public static int value(StringTokenizer in)throws Exception{
        return Integer.parseInt(in.nextToken());
        
    }
	public static void main (String[] args) throws java.lang.Exception
	{  	BufferedReader in= new BufferedReader(new InputStreamReader(System.in));  
       StringTokenizer tok= new StringTokenizer(in.readLine());
     int n=value(tok);
     int m=value(tok);
     int k=value(tok);
       int arr[][]= new int [n][m];StringTokenizer stt;
       for(int i=0;i<n;i++){
           stt=new StringTokenizer(in.readLine());
           for(int j=0;j<m;j++)
               arr[i][j]=value(stt);
                       }
    int ch[][]=new int[k][3];
    for(int i=0;i<k;i++){
        ch[i][0]=Character.getNumericValue(in.read());tok=new StringTokenizer(in.readLine());
        for(int j=1;j<3;j++)
            ch[i][j]=value(tok);  }
    StringBuilder str= new StringBuilder();
int point1 []=new int[m];
int point2 []= new int [n];
for (int i=0;i<m;i++)
    point1[i]=i;
for(int i=0;i<n;i++)
    point2[i]=i;
int x=0;
for(int i=0;i<k;i++){
    if(ch[i][0]==16&&x==0){
        str.append(arr[ch[i][1]-1][ch[i][2]-1]+"\n");


        x++;
    }
    else if(ch[i][0]==16&&x!=0){
        str.append(arr[point2[ch[i][1]-1]][point1[ch[i][2]-1]]+"\n");
        
    }
    else if(ch[i][0]==12){int temp=point1[ch[i][1]-1];
        point1[ch[i][1]-1]=point1[ch[i][2]-1];
        point1[ch[i][2]-1]=temp;x++;
    }
    else{
     int temp2=point2[ch[i][1]-1];
        point2[ch[i][1]-1]=point2[ch[i][2]-1];
        point2[ch[i][2]-1]=temp2;  x++;
        
    }
}
System.out.println(str);

	}
}