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;
typedef long long int64;
typedef unsigned long long uint64;
const double pi = acos(-1.0);
const double eps = 1e-11;
const int MAXI = 0x7fffffff;
int in() {
  int x = 0, c;
  for (; (unsigned)((c = getchar()) - '0') >= 10;) {
    if (c == '-') return -in();
    if (!~c) throw ~0;
  }
  do {
    x = (x << 3) + (x << 1) + (c - '0');
  } while ((unsigned)((c = getchar()) - '0') < 10);
  return x;
}
int main(int argc, char *argv[]) {
  long col, row, k;
  cin >> row >> col;
  k = in();
  int r[row], c[col], a[row][col];
  for (int i = (0), _b = (row); i < _b; i++)
    for (int j = (0), _b = (col); j < _b; j++) a[i][j] = in();
  for (int i = (0), _b = (row); i < _b; i++) r[i] = i;
  for (int i = (0), _b = (col); i < _b; i++) c[i] = i;
  for (int i = (0), _b = (k); i < _b; i++) {
    char s;
    int x, y;
    s = getchar();
    x = in();
    y = in();
    if (s == 'g') cout << a[r[x - 1]][c[y - 1]] << endl;
    if (s == 'r') swap(r[x - 1], r[y - 1]);
    if (s == 'c') swap(c[x - 1], c[y - 1]);
  }
  return 0;
}
