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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, k, i, j, b, c;
  char x;
  cin >> n >> m >> k;
  int a[n + 2][m + 2];
  int r[n + 2], col[m + 2];
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  for (i = 0; i < n; i++) r[i] = i;
  for (i = 0; i < m; i++) col[i] = i;
  for (i = 0; i < k; i++) {
    cin >> x >> b >> c;
    b--;
    c--;
    if (x == 'r') {
      int t = r[b];
      r[b] = r[c];
      r[c] = t;
    } else if (x == 'c') {
      int t = col[b];
      col[b] = col[c];
      col[c] = t;
    } else {
      cout << a[r[b]][col[c]] << "\n";
    }
  }
  return 0;
}
