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.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class CosmicTables {

  static BufferedReader in;
  static PrintWriter out;
  static StringTokenizer tok;

  static void solve() throws Exception {
    int n = nextInt(), m = nextInt(), k = nextInt();
    int[][] b = new int[n][m];
    for (int i = 0; i < n; i++)
      b[i] = nextIntArray(m, 0);
    int[] r = new int[n];
    for (int i = 0; i < n; i++)
      r[i] = i;
    int[] c = new int[m];
    for (int j = 0; j < m; j++)
      c[j] = j;
    for (int i = 0; i < k; i++) {
      String s = next();
      if (s.equals("r"))
        swap(r, nextInt() - 1, nextInt() - 1);
      else if (s.equals("c"))
        swap(c, nextInt() - 1, nextInt() - 1);
      else {
        int x = r[nextInt() - 1];
        int y = c[nextInt() - 1];
        out.println(b[x][y]);
      }
    }
  }

  static void swap(int[] a, int i, int j) {
    int t = a[i];
    a[i] = a[j];
    a[j] = t;
  }

  public static void main(String args[]) {
    try {
      in = new BufferedReader(new InputStreamReader(System.in));
      out = new PrintWriter(new OutputStreamWriter(System.out));
      solve();
      in.close();
      out.close();
    } catch (Throwable e) {
      e.printStackTrace();
      System.exit(1);
    }
  }

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

  static int[] nextIntArray(int len, int start) throws IOException {
    int[] a = new int[len];
    for (int i = start; i < len; i++)
      a[i] = nextInt();
    return a;
  }

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

  static long[] nextLongArray(int len, int start) throws IOException {
    long[] a = new long[len];
    for (int i = start; i < len; i++)
      a[i] = nextLong();
    return a;
  }

  static String next() throws IOException {
    while (tok == null || !tok.hasMoreTokens()) {
      tok = new StringTokenizer(in.readLine());
    }
    return tok.nextToken();
  }

}
