題幹：

You are given a square matrix consisting of n rows and n columns. We assume that the rows are numbered from 1 to n from top to bottom and the columns are numbered from 1to n from left to right. Some cells (n - 1 cells in total) of the the matrix are filled with ones, the remaining cells are filled with zeros. We can apply the following operations to the matrix:

1. Swap i-th and j-th rows of the matrix;
2. Swap i-th and j-th columns of the matrix.

You are asked to transform the matrix into a special form using these operations. In that special form all the ones must be in the cells that lie below the main diagonal. Cell of the matrix, which is located on the intersection of the i

-th row and of the j-th column, lies below the main diagonal if i > j.

Input

The first line contains an integer n (2 ≤ n ≤ 1000) — the number of rows and columns. Then follow n - 1 lines that contain one's positions, one per line. Each position is described by two integers x

k, yk (1 ≤ xk, yk ≤ n), separated by a space. A pair (xk, yk) means that the cell, which is located on the intersection of the xk-th row and of the yk-th column, contains one.

It is guaranteed that all positions are distinct.

Output

Print the description of your actions. These actions should transform the matrix to the described special form.

In the first line you should print a non-negative integer m (m ≤ 105) — the number of actions. In each of the next m lines print three space-separated integers t, i, j(1 ≤ t ≤ 2, 1 ≤ i, j ≤ n, i ≠ j), where t = 1 if you want to swap rows, t = 2 if you want to swap columns, and i and j denote the numbers of rows or columns respectively.

Please note, that you do not need to minimize the number of operations, but their number should not exceed 105. If there are several solutions, you may print any of them.

Examples

Input

2
1 2

Output

2
2 1 2
1 1 2

Input

3
3 1
1 3

Output

3
2 2 3
1 1 3
1 1 2

Input

3
2 1
3 2

Output

0

題目大意：

給你一個n*n的矩陣並且其中有n-1個位置是1（其他位置是0）

每次可以交換其中任意兩行或者兩列

求在10^5個交換內把所有的1都交換到對角線以下（橫座標大於縱座標）的方法

答案可以有很多，寫出任意一個即可。（並且某次交換中 行==列  是不允許的）

解題報告：

   其實這道題想出來就不難了，，，程式碼也不算難實現，，坑也不算多。。（問題就是想不到啊還是做題少了）

   思路就是我們因為有n-1個1，所以肯定存在一列裡面全為0，此時把這一列移到最右邊。然後再找一行有1的，把他和最後一行交換。（也可以  如果最後一行全為0  ，才執行第二個操作）

此時，我們就可以把問題變成在前（n-1）*（n-1）的矩陣中把最多n-2個點移到左下角的問題

   有個小細節，因為行列相等不算交換，，所以迴圈中就不能加for(i=1;i<=n;i++)中的等號了。

AC程式碼：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int maze[1005][1005];
int x[MAX],y[MAX];
int tot;
struct Node {
	int t,a,b;
	Node(){}
	Node(int t,int a,int b):t(t),a(a),b(b){}
} node[MAX];
void dfs(int n) {
	if(n == 1) return ;
	for(int j = 1; j<n; j++) {//列舉每一列 
		if(y[j] == 0) {
			for(int i = 1; i<=n; i++) swap(maze[i][j],maze[i][n]);
			swap(y[j],y[n]);
			node[++tot] = Node(2,j,n); 
			break;
		}
	}
	for(int i = 1; i<n; i++) {//列舉每一行 
		if(x[i] != 0) {
			for(int j = 1; j<=n; j++) swap(maze[i][j],maze[n][j]);
			swap(x[i],x[n]);
			node[++tot] = Node(1,i,n);
			break;
		}
	}
	for(int j = 1; j<=n-1; j++) {
		if(maze[n][j] == 1) y[j]--;
	}
	dfs(n-1);
}
int main()
{
	int all;
	cin>>all;
	for(int i = 1; i<=all-1; i++) {
		int a,b;
		scanf("%d%d",&a,&b);
		maze[a][b]=1;
		x[a]++;y[b]++;
	}
	dfs(all);
	printf("%d\n",tot);
	for(int i = 1; i<=tot; i++) printf("%d %d %d\n",node[i].t,node[i].a,node[i].b);
	return 0 ;
 }