Garland（CodeForces

C. Garland

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Once at New Year Dima had a dream in which he was presented a fairy garland. A garland is a set of lamps, some pairs of which are connected by wires. Dima remembered that each two lamps in the garland were connected directly or indirectly via some wires. Furthermore, the number of wires was exactly one less than the number of lamps.

There was something unusual about the garland. Each lamp had its own brightness which depended on the temperature of the lamp. Temperatures could be positive, negative or zero. Dima has two friends, so he decided to share the garland with them. He wants to cut two different wires so that the garland breaks up into three parts. Each part of the garland should shine equally, i. e. the sums of lamps' temperatures should be equal in each of the parts. Of course, each of the parts should be non-empty, i. e. each part should contain at least one lamp.

 

Help Dima to find a suitable way to cut the garland, or determine that this is impossible.

While examining the garland, Dima lifted it up holding by one of the lamps. Thus, each of the lamps, except the one he is holding by, is now hanging on some wire. So, you should print two lamp ids as the answer which denote that Dima should cut the wires these lamps are hanging on. Of course, the lamp Dima is holding the garland by can't be included in the answer.

Input

The first line contains single integer n (3 ≤ n ≤ 106) — the number of lamps in the garland.

Then n lines follow. The i-th of them contain the information about the i-th lamp: the number lamp ai, it is hanging on (and 0, if is there is no such lamp), and its temperature ti ( - 100 ≤ ti ≤ 100). The lamps are numbered from 1 to n.

Output

If there is no solution, print -1.

Otherwise print two integers — the indexes of the lamps which mean Dima should cut the wires they are hanging on. If there are multiple answers, print any of them.

Examples

input

6
2 4
0 5
4 2
2 1
1 1
4 2

output

1 4

input

6
2 4
0 6
4 2
2 1
1 1
4 2

output

-1

Note

The garland and cuts scheme for the first example:

 

 

题意理解：这道题，根据他输入的顺序，每个数都会有一个编号，每行输入，第一个数表示的是他所指向的灯泡编号，第二个便是他自己的亮度值。此题就是要剪两刀，剪成三段一样亮度值的东西。

解题思路：主要就是先把每个点发展下去所能有的亮度值，标记出来，遇到是总亮度值除以三就可以了。然后把这个点的值变为零。还有就是要注意一下，不能把最开始的那个根节点减掉。

#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>#include<stack>#include<vector>#include<map>#include<math.h>#include<set>#include<string>using namespace std;int pre[1000010];int val[1000010];int sum_[1000010];vector<int>s[1000010]; int sum=0;int cnt=0;int ans1=0;int ans2=0;void dfs(int numb){    sum_[numb]=val[numb];    for(int i=0;i<s[numb].size();i++)    {        if(s[numb][i]!=pre[numb])        {            dfs(s[numb][i]);            sum_[numb]=sum_[numb]+sum_[s[numb][i]];   //核心标记        }    }    if(sum_[numb]==sum/3)    {       if(pre[numb]!=0)   //不能是根节点       {          if(cnt==0)        {             ans1=numb;             sum_[numb]=0;             cnt++;        }        else if(cnt==1)        {            ans2=numb;            sum_[numb]=0;            cnt++;        }       }    }}int main(){      int n;      int root;      int num=1;      scanf("%d",&n);      while(n--)      {          int x,y;          scanf("%d%d",&x,&y);          if(x==0)          {               root=num;               val[num]=y;               sum+=y;               pre[num]=x;          }          else          {               pre[num]=x;               val[num]=y;               sum+=y;               s[num].push_back(x);               s[x].push_back(num);          }          num++;      }      if(sum%3!=0)      {          printf("-1\n");      }      else      {          dfs(root);          if(cnt>=2)          {              printf("%d %d\n",ans1,ans2);          }          else          {              printf("-1\n");          }      }}