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.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Locale;
import java.util.StringTokenizer;

public class Problem implements Runnable {
    public static void main(String[] args) {
        new Problem().run();
    }
    
    final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;
    
    public String nextToken() throws IOException {
        while (in == null || !in.hasMoreTokens())
            in = new StringTokenizer(br.readLine());
        return in.nextToken();
    }
    
    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
    
    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
    
    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
    
    void solve() throws IOException {
        int n = nextInt();
        int m = nextInt();
        int k = nextInt();
        int[][] a = new int[n][m];
        int[] r = new int[n];
        int[] c = new int[m];
        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 < n; i++)
            for (int j = 0; j < m; j++)
                a[i][j] = nextInt();
        for (int i = 0; i < k; i++) {
            char q = nextToken().charAt(0);
            int x = nextInt() - 1;
            int y = nextInt() - 1;
            if (q == 'c') {
                int tmp = c[x];
                c[x] = c[y];
                c[y] = tmp;
            }
            else if (q == 'r') {
                int tmp = r[x];
                r[x] = r[y];
                r[y] = tmp;
            }
            else
                out.println(a[r[x]][c[y]]);
        }
    }
    
    public void run() {
        try {
            long startTime = System.currentTimeMillis();       
            Locale.setDefault(Locale.US);       
            if (!ONLINE_JUDGE) {
                br = new BufferedReader(new FileReader("input.txt"));
                out = new PrintWriter("output.txt");
                solve();
            } else {
                br = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(System.out);
                solve();
            }
            out.close();
            br.close();    
            long endTime = System.currentTimeMillis();
            long freeMemory = Runtime.getRuntime().freeMemory();
            long totalMemory = Runtime.getRuntime().totalMemory();
            System.err.printf("Time = %.3f s\n", (endTime - startTime) / 1000.0);
            System.err.printf("Memory = %.3f MB\n", (totalMemory - freeMemory) / (float)(1 << 20));
            
        } catch (Throwable t) {
            t.printStackTrace(System.err);
            System.exit(-1);
        }
    }
}