CodeForces - 767C Garland 树的遍历

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:

树的后序遍历，根节点的值为子树权值和。当和为sum/3时，节点值赋0，返给父节点（相当于剪掉）。当找到两个非根结点的sum/3权值和时，即为答案。注意必须是最先找到的前两个，虽然判断了不是根节点，但是会WA在第九点...

#include<stdio.h>
#include<vector>
using namespace std;

int b[],w[],s[];
vector<int> v[];
int root,co,c1,c2;

int dfs(int f)
{
    int i;
    for(i=;i<v[f].size();i++){
        if(b[v[f][i]]==){
            b[v[f][i]]=;
            s[f]+=dfs(v[f][i]);
        }
    }
    s[f]+=w[f];
    if(s[f]==co/&&f!=root){
        if(c1==) c1=f;
        else if(c2==) c2=f;   //不加c2==0WA。。
        return ;
    }
    return s[f];
}

int main()
{
    int n,x,y,i;
    scanf("%d",&n);
    co=;
    for(i=;i<=n;i++){
        scanf("%d%d",&x,&y);
        if(x==) root=i;
        else{
            v[i].push_back(x);
            v[x].push_back(i);
        }
        w[i]=y;
        co+=y;
    }
    if(co%) printf("-1\n");
    else{
        c1=;c2=;
        b[root]=;
        dfs(root);
        if(c1!=&&c2!=&&c1!=c2) printf("%d %d\n",c1,c2);
        else printf("-1\n");
    }
    return ;
}