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;
const int dr[] = {0, -1, 0, 1, -1, 1, 1, -1};
const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1};
const double eps = 1e-9;
template <class T>
int len(const T &c) {
  return (int)c.size();
}
template <class T>
void cmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
void cmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T, class U>
T cast(U first) {
  T second;
  ostringstream a;
  a << first;
  istringstream b(a.str());
  b >> second;
  return second;
}
vector<string> split(string s, string first = " ") {
  vector<string> res;
  for (int i(0), _n(len(s)); i < _n; ++i) {
    string a;
    while (i < (int)s.size() && first.find(s[i]) == string::npos) a += s[i++];
    if (!a.empty()) res.push_back((a));
  }
  return res;
}
template <class T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <class T>
inline T Pow(T b, T p) {
  T ret = 1;
  for (int i(0), _n(p); i < _n; ++i) {
    ret *= b;
  }
  return ret;
}
int n, m, k, t1, t2, t3;
string s;
vector<int> vr[1001];
vector<int> rc;
vector<int> cc;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> k;
  for (int i(0), _n(1001); i < _n; ++i) rc.push_back(i), cc.push_back(i);
  for (int i(0), _n(n); i < _n; ++i)
    for (int j(0), _n(m); j < _n; ++j) {
      cin >> t1;
      vr[i].push_back(t1);
    }
  for (int i(0), _n(k); i < _n; ++i) {
    cin >> s >> t1 >> t2;
    t1--;
    t2--;
    if (s == "r") {
      swap(rc[t1], rc[t2]);
    } else if (s == "c") {
      swap(cc[t1], cc[t2]);
    } else {
      cout << vr[rc[t1]][cc[t2]] << endl;
    }
  }
}
