Description：

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.

这道题就是一道简单的模拟题，但是直接暴力的话可能会超时，昨天可能是数据水暴力不超时，可以只交换下标这样复杂度会降低不少。

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stdlib.h>
#include<stack>
#include<vector>
#include<queue>
const int INF = 0x3f3f3f3f;
using namespace std;
int a[1001][1001];
int h[1001],l[1001];
int main()
{
    int n,m,k,i,j;
    char op;
    int pos1,pos2;
    int temp;
    scanf("%d %d %d",&n,&m,&k);
    for(i=0;i<1001;i++)
    {
        h[i]=i;
        l[i]=i;
    }
    for(i=0;i<n;i++)
    {
        for(j=0;j<m;j++)
        {
            scanf("%d",&a[i][j]);
        }
    }
    for(i=0;i<k;i++)
    {
        getchar();
        scanf("%c %d %d",&op,&pos1,&pos2);
        pos1--;
        pos2--;
        if(op=='g')
        {
            printf("%d\n",a[h[pos1]][l[pos2]]);
        }
        else if(op=='c')
        {
            swap(l[pos1],l[pos2]);
        }
        else if(op=='r')
        {
             swap(h[pos1],h[pos2]);
        }
    }
    return 0;
}