The board has got a painted tree graph, consisting of n nodes. Let us remind you that a non-directed graph is called a tree if it is connected and doesn't contain any cycles.
Each node of the graph is painted black or white in such a manner that there aren't two nodes of the same color, connected by an edge. Each edge contains its value written on it as a non-negative integer.
A bad boy Vasya came up to the board and wrote number sv near each node v — the sum of values of all edges that are incident to this node. Then Vasya removed the edges and their values from the board.
Your task is to restore the original tree by the node colors and numbers sv.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 105) — the number of nodes in the tree. Next n lines contain pairs of space-separated integers ci, si (0 ≤ ci ≤ 1, 0 ≤ si ≤ 109), where ci stands for the color of the i-th vertex (0 is for white, 1 is for black), and si represents the sum of values of the edges that are incident to the i-th vertex of the tree that is painted on the board.
Output
Print the description of n - 1 edges of the tree graph. Each description is a group of three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 0 ≤ wi ≤ 109), where vi and ui — are the numbers of the nodes that are connected by the i-th edge, and wi is its value. Note that the following condition must fulfill cvi ≠ cui.
It is guaranteed that for any input data there exists at least one graph that meets these data. If there are multiple solutions, print any of them. You are allowed to print the edges in any order. As you print the numbers, separate them with spaces.
Examples
Input
3
1 3
1 2
0 5
Output
3 1 3
3 2 2
Input
6
1 0
0 3
1 8
0 2
0 3
0 0
Output
2 3 3
5 3 3
4 3 2
1 6 0
2 1 0题解： 我们先把两种颜色的点放在两个优先队列中，每次选出价值最大的，然后在把价值大的点，重新放进队列，此时，该点的价值变小了，还要注意的是，若两个点价值相等，并且已经出现空队列了，那就把原先的点放进去，并且价值为0，为什么呢？
因为可能会出现 比如 白色 还有2个点 价值为0 而黑色的点都pop掉了 所以 保证每个队列都要有元素，刚开始写的时候没考虑，错在了 第9个 纳闷为什么没有报RT
#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
const int N=1e5+10;
struct node{
int id,val;
node(){}
node(int id,int val):id(id),val(val){}
bool operator < (const node &x)const {
return val < x.val;
}
};
struct edge{
int u,v,d;
}e[N];
int n;
int main()
{
while(~scanf("%d", &n))
{
priority_queue<node> p1,p2;
node now1,now2;
int op,val;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&op,&val);
if(op==0)
p1.push(node(i,val));
else
p2.push(node(i,val));
}
int len=0;
for(int i=1;i<n;i++)
{
now1=p1.top();p1.pop();
now2=p2.top();p2.pop();
e[++len].u=now1.id;
e[len].v=now2.id;
e[len].d=min(now1.val,now2.val);
if(now1.val>now2.val)
p1.push(node(now1.id,now1.val-now2.val));
else if(now1.val<now2.val)
p2.push(node(now2.id,now2.val-now1.val));
else
{
if(p1.empty())
p1.push(node(now1.id,0));
else
p2.push(node(now2.id,0));
}
}
for(int i=1;i<n;i++)
printf("%d %d %d\n",e[i].u,e[i].v,e[i].d);
}
return 0;
}