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>
#pragma GCC optimize("O2")
using namespace std;
const int INF = 1 << 29;
const int MOD = 1e9 + 7;
bool isPowerOfTwo(long long int x) { return x && (!(x & (x - 1))); }
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
const int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[] = {0, -1, 0, 1, 1, -1, -1, 1};
int main() {
  fastio();
  int tc = 1;
  while (tc--) {
    int n, m, k;
    cin >> n >> m >> k;
    int a[n][m];
    for (int i = 0; i < (n); i++) {
      for (int j = 0; j < (m); j++) cin >> a[i][j];
    }
    map<int, int> r, c, r2, c2;
    for (int i = 0; i < (n); i++) r[i] = i;
    for (int i = 0; i < (m); i++) c[i] = i;
    while (k--) {
      char c1;
      cin >> c1;
      int x, y;
      cin >> x >> y;
      x--, y--;
      if (c1 == 'g') {
        cout << a[r[x]][c[y]];
        cout << "\n";
        ;
      } else if (c1 == 'c') {
        int uk = c[x], uk1 = c[y];
        c[x] = uk1;
        c[y] = uk;
      } else {
        int uk = r[x], uk1 = r[y];
        r[x] = uk1;
        r[y] = uk;
      }
    }
  }
  return 0;
}
