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;
long long power(long x, long n) {
  if (n == 0) return 1;
  if (n == 1) return x;
  long n2 = n / 2;
  long long po = power(x, n2);
  if (n % 2) return po * po * x;
  return po * po;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
class BCosmicTables {
 public:
  void solve(std::istream &cin, std::ostream &cout) {
    int n, m, c, r, q, x, y;
    cin >> n >> m >> q;
    int A[n][m];
    for (r = 0; r < n; r++) {
      for (c = 0; c < m; c++) {
        cin >> A[r][c];
      }
    }
    map<int, int> R, C;
    for (r = 0; r < n; r++) R[r] = r;
    for (c = 0; c < m; c++) C[c] = c;
    char ch;
    while (q--) {
      cin >> ch >> x >> y;
      if (ch == 'r') {
        r = R[x - 1];
        c = R[y - 1];
        R[y - 1] = r;
        R[x - 1] = c;
      } else if (ch == 'c') {
        r = C[x - 1];
        c = C[y - 1];
        C[y - 1] = r;
        C[x - 1] = c;
      } else {
        r = R[x - 1];
        c = C[y - 1];
        cout << A[r][c] << '\n';
      }
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  BCosmicTables solver;
  std::istream &in(std::cin);
  std::ostream &out(std::cout);
  solver.solve(in, out);
  return 0;
}
