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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.util.Scanner;

public class B {
    class FastIO {
        
        int nextInt() throws IOException {
            return Integer.valueOf(next());
        }
        
        long nextLong() throws IOException {
            return Long.valueOf(next());
        }
        
        double nextDouble() throws IOException{
            return Double.valueOf(next());
        }

        String next() throws IOException {
            StringBuilder sb = new StringBuilder();
            char c = skipChar();
            while (Character.isWhitespace(c))
                c = skipChar();
            while (!Character.isWhitespace(c)) {
                sb.append(c);
                c = getChar();
            }
            return sb.toString();
        }

        char skipChar() throws IOException {
            int tmp = bfr.read();
            if (tmp == -1)
                throw new InputMismatchException();
            return (char) tmp;
        }

        char getChar() throws IOException {
            return (char) bfr.read();
        }

        void end() {
            out.flush();
            out.close();
        }

        BufferedReader bfr = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    }

    static int maxn=1010;
    static int map[][]=new int[maxn][maxn];
    static int N,M,K;
    static int temp[]=new int[maxn];
    static int hang[]=new int[maxn];
    static int lie[]=new int[maxn];
    static void run1(int x,int y)//hang
    {
        
        int temp=hang[x];
        hang[x]=hang[y];
        hang[y]=temp;
        
    }
    
    static void run2(int x,int y)
    {
        int temp=lie[x];
        lie[x]=lie[y];
        lie[y]=temp;
    }
    
     void run() throws IOException {
        
//      Scanner in=new Scanner(System.in);
        FastIO in=new FastIO();
        N=in.nextInt();
        M=in.nextInt();
        K=in.nextInt();
        for(int i=1;i<=N;i++)
            hang[i]=i;
        for(int i=1;i<=M;i++)
            lie[i]=i;
        
        for(int i=1;i<=N;i++)
        {
            for(int j=1;j<=M;j++)
            {
                map[i][j]=in.nextInt();
            }
        }
        
        while(K--!=0)
        {
            String inc=in.next();
            int x=in.nextInt();
            int y=in.nextInt();
            
            if(inc.equals("r"))
            {
                run1(x,y);
                
            }else if(inc.equals("c"))
            {
                run2(x,y);
            }else if(inc.equals("g"))
            {
                System.out.println(map[hang[x]][lie[y]]);
            }
        }
    }
     public static void main(String[] args) throws IOException {
        new B().run();
    }
}
