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>
const int buf_size = 4096;
unsigned char buf[4096];
int buf_len = 0, buf_pos = 0;
inline bool isEof() {
  if (buf_pos == buf_len) {
    buf_pos = 0, buf_len = fread(buf, 1, buf_size, stdin);
    if (buf_pos == buf_len) return 1;
  }
  return 0;
}
inline int getChar() { return isEof() ? -1 : buf[buf_pos++]; }
inline int readChar() {
  int c = getChar();
  while (c != -1 && c <= 32) c = getChar();
  return c;
}
inline int readInt() {
  int s = 1, c = readChar();
  int x = 0;
  if (c == '-')
    s = -1, c = getChar();
  else if (c == '+')
    c = getChar();
  while ('0' <= c && c <= '9') x = x * 10 + c - '0', c = getChar();
  return s == 1 ? x : -x;
}
int write_buf_pos = 0;
char write_buf[4096];
inline void writeChar(int x) {
  if (write_buf_pos == buf_size)
    fwrite(write_buf, 1, buf_size, stdout), write_buf_pos = 0;
  write_buf[write_buf_pos++] = x;
}
inline void flush() {
  if (write_buf_pos)
    fwrite(write_buf, 1, write_buf_pos, stdout), write_buf_pos = 0;
}
inline void writeInt(int x, char end) {
  int output_len = -1;
  if (x < 0) writeChar('-'), x = -x;
  char s[24];
  int n = 0;
  while (x || !n) s[n++] = '0' + x % 10, x /= 10;
  while (n < output_len) s[n++] = '0';
  while (n--) writeChar(s[n]);
  if (end) writeChar(end);
}
int n, m, k, t, a[1024][1024], x, y, cols[1024], rows[1024];
char c;
int main() {
  n = readInt(), m = readInt(), k = readInt();
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) a[i][j] = readInt();
  for (int i = 1; i <= n; ++i) rows[i] = i;
  for (int i = 1; i <= m; ++i) cols[i] = i;
  while (k--) {
    c = readChar(), x = readInt(), y = readInt();
    if (c == 'c')
      t = cols[x], cols[x] = cols[y], cols[y] = t;
    else if (c == 'r')
      t = rows[x], rows[x] = rows[y], rows[y] = t;
    else
      writeInt(a[rows[x]][cols[y]], '\n');
  }
  flush();
}
