[贪心]cf225bMultitasking

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本文探讨了Iahub如何通过多任务处理优化数组排序效率。通过选择特定的数组索引来交换元素，实现数组的升序或降序排列。详细介绍了排序策略及输出排序所需的操作次数。

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Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.

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

Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most $\text{[math]}$ (at most $\text{[math]}$ 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 contains m integers separated by a space, representing the i-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 $\text{[math]}$ ). Each of the next p lines must contain two distinct integers i 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 0
1 3 2 5 4
1 4 3 2 5

Output

Input

Output

1
2 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 $\text{[math]}$ )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);
			else
				printf("%d %d",i+1,i);
			if (k!=m-1 || i!=m-k)
				printf("\n"); }
	}
	return 0;
}