Codeforces Round #398(Div. 2)C. Garland【Dfs+思维】

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本文探讨了一种特殊的树形结构问题，旨在通过切断两根树枝的方式，将树分为三个部分，使每个部分的权值之和相等。文章详细介绍了问题背景、解决思路及AC代码实现。

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.

$\text{[math]}$

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 a_i, it is hanging on (and 0, if is there is no such lamp), and its temperature t_i ( - 100 ≤ t_i ≤ 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

Output

1 4

Input

Output

-1

Note

The garland and cuts scheme for the first example:

$\text{[math]}$

题目大意：

给你一颗树，其中包含N个点，让你将这颗树剪去两条边，使得变成三部分，要求三部分的权值和相等。

思路：

1、我们设定sum【i】表示以i作为根的子树权值和。

那么过程维护，sum【i】+=sum【v】；

2、如果全部点的权值和不是3的倍数，那么显然无解，否则考虑：

①遇到了一个点u，此时sum【u】=tot/3；那么这个点就可以将其拆除出来。

②将一个点拆出来的同时，设定sum【u】=0；回溯，继续找下一个tot/3

③如果过程中遇到了两个sum【u】=tot；此时存在解。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
vector<int >mp[1000070];
int f[1000050];
int val[1000050];
int sum[1000060];
int tot,ok;
int ans1,ans2;
void Dfs(int u,int from)
{
    sum[u]=val[u];
    for(int i=0;i<mp[u].size();i++)
    {
        int v=mp[u][i];
        if(v==from)continue;
        Dfs(v,u);
        sum[u]+=sum[v];
    }
    if(sum[u]==tot)
    {
        if(from==-1)return ;
        if(ans1==-1)
        {
            ans1=u;
            sum[u]=0;
        }
        else if(ans2==-1)
        {
            ans2=u;
            sum[u]=0;
        }
    }
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<=n;i++)mp[i].clear();
        int root;
        tot=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&f[i],&val[i]);
            tot+=val[i];
            if(f[i]==0)
            {
                root=i;
                continue;
            }
            mp[i].push_back(f[i]);
            mp[f[i]].push_back(i);
        }
        if(tot%3!=0)
        {
            printf("-1\n");
            continue;
        }
        ans1=-1;
        ans2=-1;
        tot/=3;
        ok=0;
        Dfs(root,-1);
        if(ans1==-1||ans2==-1)printf("-1\n");
        else printf("%d %d\n",ans1,ans2);
    }
}