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.util.*;
import java.io.*;

public class CodeForces {
    static int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
    static int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
    static final int INF = (int) 1e9 + 10;
    
    public static void main(String[] args) throws IOException {
        in.init(System.in);
        
        int n = in.i(), m = in.i(), q = in.i();
        int arr[][] = new int[n][m];
        int mapR[] = new int[n+1];
        int mapC[] = new int[m+1];
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                arr[i][j] =  in.i();
                mapC[j+1] = j+1;
            }
            
            mapR[i+1] = i+1;
        }
        
        StringBuilder ans = new StringBuilder();
        
        for (int i = 0; i < q; i++) {
            char ch = in.n().charAt(0);
            int a = in.i(), b = in.i();
            
            if(ch == 'c'){
                int temp = mapC[a];
                mapC[a] = mapC[b];
                mapC[b] = temp;
            }
            
            else if(ch == 'r'){
                int temp = mapR[a];
                mapR[a] = mapR[b];
                mapR[b] = temp;
            }
            
            else{
                b = mapC[b];
                a = mapR[a];
                
                ans.append(arr[a-1][b-1]).append('\n');
            }
        }
        
        System.out.println(ans);
    }
}

class Queue{
    int first = -1, last = -1, size = 0;
    long arr[];
    
    Queue(int n){
        arr = new long[n+10];
    }
    
    long pop(){
        size--;
        return arr[++first];
    }
    
    void push(long x){
        size++;
        arr[++last] = x;
    }
    
    boolean isEmpty(){
        return first == last;
    }
}
class Pair implements Comparable<Pair> {

    int x;
    int y;

    Pair(int a, int b) {
        x = a;
        y = b;
    }

    @Override
    public int compareTo(Pair t) {
        return x - t.x;
    }

    @Override
    public int hashCode() {
        return 1;
    }

    @Override
    public boolean equals(Object o) {
        return ((Pair)o).x == x && ((Pair)o).y == y;
    }
    
    
    
    
}

class in {

    static BufferedReader reader;
    static StringTokenizer tokenizer;

    public static int pars(String x) {
        int num = 0;
        int i = 0;
        if (x.charAt(0) == '-') {
            i = 1;
        }
        for (; i < x.length(); i++) {
            num = num * 10 + (x.charAt(i) - '0');
        }

        if (x.charAt(0) == '-') {
            return -num;
        }

        return num;
    }

    static void init(InputStream input) {
        reader = new BufferedReader(
                new InputStreamReader(input));
        tokenizer = new StringTokenizer("");
    }

    static void init(FileReader input) {
        reader = new BufferedReader(input);
        tokenizer = new StringTokenizer("");
    }

    static String n() throws IOException {
        while (!tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(
                    reader.readLine());
        }
        return tokenizer.nextToken();
    }

    static int i() throws IOException {
        return pars(n());
    }

    static long l() throws IOException {
        return Long.parseLong(n());
    }

    static double d() throws IOException {
        return Double.parseDouble(n());
    }
}