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.awt.Point;
import java.io.*;
import java.util.*;
import java.math.BigInteger;

public class CosmicTables {
    static class Solver {
        public void solve(MyReader in, PrintWriter out) {
            int n= in.nextInt();
            int m = in.nextInt();

            int k = in.nextInt();

            int[] rows = new int[n];
            int[] cols = new int[m];

            for(int i = 0;i<n;i++)rows[i] = i;

            for (int i = 0; i < m; i++) {
                cols[i] = i;
            }

            int[][] mat = new int[n][m];

            for(int i = 0;i<n;i++){
                for (int j = 0; j < m; j++) {
                    mat[i][j] = in.nextInt();
                }
            }

            for (int i = 0; i < k; i++) {
                char c = in.next().charAt(0);
                int x = in.nextInt()-1;
                int y = in.nextInt()-1;

                if(c=='g'){
                    out.println(mat[rows[x]][cols[y]]);
                }
                else if(c == 'r'){
                    swap(rows,x,y);
                }
                else{
                    swap(cols,x,y);
                }
            }



        }


    }

    public static void main(String[] args) {
        MyReader mr = new MyReader();
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        new Solver().solve(mr, out);
        out.close();
    }

    //    static int[][] dir = {{0,1}, {1,0},{-1,0}, {0,-1}};
//    static int[][] dir = {{0,1}, {1,0},{-1,0}, {0,-1}, {-1,-1},{-1,1}, {1,1}, {1,-1}};
    static class MyReader {
        BufferedReader br;
        StringTokenizer st;

        MyReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }

            return st.nextToken();
        }

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

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

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

        String nextLine() {
            String res = "";
            try {
                res = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }

            return res;
        }

        int[] nextIntArray(int n) {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }

            return arr;
        }

        Integer[] nextIntegerArray(int n) {
            Integer[] arr = new Integer[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }

            return arr;
        }

        Long[] nextLongArray(int n) {
            Long[] arr = new Long[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextLong();
            }

            return arr;
        }

        String[] nextStringArray(int n) {
            String[] arr = new String[n];
            for (int i = 0; i < n; i++) {
                arr[i] = next();
            }
            return arr;
        }
    }

    static void swap(int[] arr, int i, int j) {
        int tmp = arr[i];
        arr[i] = arr[j];
        arr[j] = tmp;
    }

}


