[贪心]cf225bMultitasking

B. Multitasking

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sortn arrays simultaneously, each array consisting ofm integers.

Iahub can choose a pair of distinct indices i andj (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positionsi and j are swapped only if the value at positioni is strictly greater than the value at positionj.

Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of then arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most (at most pairs). Help Iahub, find any suitable array.

Input

The first line contains three integers n(1 ≤  n ≤ 1000), m(1 ≤ m ≤  100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines containsm integers separated by a space, representing thei-th array. For each element x of the array i, 1 ≤ x ≤ 10⁶ holds.

Output

On the first line of the output print an integer p, the size of the array (p can be at most). Each of the nextp lines must contain two distinct integersi and j(1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.

If there are multiple correct answers, you can print any.

Sample test(s)

Input

2 5 01 3 2 5 41 4 3 2 5

Output

32 42 34 5

Input

3 2 11 22 33 4

Output

12 1

Note

Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become[1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become[1, 2, 3, 4, 5] and [1, 2, 3, 4, 5].

(p can be at most )是最重要的条件，这正是所有的元素两两交换的总次数，而题中告诉交换到“好”的，就自动固定，不会再动。因此我们输出两两交换就行了。注意有先后顺序。

#include <cstdio>#include <cstring>int cnt[200];int num[200];int tmp[200];void swap(int& a,int& b){int t = a;a = b;b = t;}int main(){//freopen("b.in","r",stdin);//freopen("b.out","w",stdout);int t;int n,m;scanf("%d%d%d",&n,&m,&t);printf("%d\n",m*(m-1)/2);for (int k=1;k<m;k++){for (int i=1;i<=m-k;i++){if (t == 0)printf("%d %d",i,i+1);elseprintf("%d %d",i+1,i);if (k!=m-1 || i!=m-k)printf("\n"); }}return 0;}