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.*;
import java.util.*;
public class Main {
    public static void main(String[] args) throws Exception {      
        Reader.init(System.in);        
        int n = Reader.nextInt() ; 
        int m = Reader.nextInt() ; 
        int k = Reader.nextInt() ; 
        StringBuilder res = new StringBuilder ()  ;
        int [][] a = new int [n][m] ; 
        int [] r = new int [n] ; 
        int [] c = new int [m] ; 
        for (int i=0 ; i<n ; i++) 
            for (int j=0 ; j<m ; j++)
                a[i][j]=Reader.nextInt() ;
        for (int i=0 ; i<n ; i++)
            r[i]=i ; 
        for (int i=0 ; i<m ; i++)
            c[i]=i ; 
        for (int i=0 ; i<k ; i++) {
            String s = Reader.next() ;
            int x = Reader.nextInt()-1 ;
            int y = Reader.nextInt()-1 ; 
            if (s.equals("r")) {
                int temp = r[x] ; 
                r[x]=r[y];
                r[y]=temp ; 
            } else if (s.equals("c")) {
                int temp = c[x];
                c[x]=c[y];
                c[y]=temp; 
            } else {
                res.append(a[r[x]][c[y]]+"\n");
            }
        }
        System.out.print(res);
        
    }     
    public static int BS(int[] a, int key, int num) {
        int low = 0;
        int high = a.length - 1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (a[mid] == key && mid != num) {
                return mid;
            } else if (a[mid] < key) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return -1;
    }

    public static boolean isSorted(int[] a) {

        int i;
        for (i = 0; i < a.length - 1; i++) {
            if (a[i] < a[i + 1]) {
                continue;
            } else {
                return false;
            }
        }
        return true;
    }

    public static void fill(int[] a) throws Exception {
        for (int i = 0; i < a.length; i++) {
            a[i] = Reader.nextInt();
        }
    }

    static void s(int[] cc) {
        for (int e = 0; e < cc.length; e++) {
            for (int q = e + 1; q < cc.length; q++) {
                if (cc[e] < cc[q]) {
                    int g = cc[e];
                    cc[e] = cc[q];
                    cc[q] = g;
                }
            }
        }

    }

    static boolean isPalindrome(String s) {
        int n = s.length();
        for (int i = 0; i < (n / 2) + 1; ++i) {
            if (s.charAt(i) != s.charAt(n - i - 1)) {
                return false;
            }
        }

        return true;
    }
}
class MyPair
{
     String s ;
     int x;
     int y;

    public MyPair(String s,int x, int y)
    {
        this.s = s;
        this.x = x;
        this.y = y;
    }

}
class Reader {

    static BufferedReader reader;
    static StringTokenizer tokenizer;

    /**
     * call this method to initialize reader for InputStream
     */
    static void init(InputStream input) {
        reader = new BufferedReader(
                new InputStreamReader(input));
        tokenizer = new StringTokenizer("");
    }

    /**
     * get next word
     */
    static String next() throws IOException {
        while (!tokenizer.hasMoreTokens()) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                    reader.readLine());
        }
        return tokenizer.nextToken();
    }

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

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

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