Codeforces Round #398 (Div. 2) C. Garland 后序遍历树

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

62 40 54 22 11 14 2

output

1 4

input

62 40 64 22 11 14 2

output

-1

Note

The garland and cuts scheme for the first example:

对树后序遍历一遍，s【i】里存的是结点i上所有子树结点权值和（包括这个结点，但不包括那些已经切去的结点）

从叶子往上遍历，当s【i】等于sum/3时，就在这个点切，当切去两个部分之后就不切了，最后看看剩下的部分等不等于sum/3就行了

#include<stdio.h>#include<string.h>#include<math.h>#include<string>#include<algorithm>#include<queue>#include<vector>#include<map>#include<set>#define eps 1e-9#define PI 3.141592653589793#define bs 1000000007#define bsize 256#define MEM(a) memset(a,0,sizeof(a))typedef long long ll;using namespace std;int n,t[1000005],s[1000005],sum=0,flog=0,beginn,ans[5];vector<int>w[1000005];int dfs(int root){int index;for(int i=0;i<w[root].size();i++){index=w[root].at(i);dfs(index);s[root]+=s[index];}if(s[root]==sum&&root!=beginn&&flog<2){ans[flog++]=root;s[root]=0;return 1;}return 0;}int main(){int i,a;MEM(s);scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&a);if(a==0)beginn=i;elsew[a].push_back(i);scanf("%d",&t[i]);s[i]=t[i];sum+=t[i];}if(sum%3)printf("-1\n");else{sum/=3;dfs(beginn);if(flog==2&&s[beginn]==sum)printf("%d %d\n",ans[0],ans[1]);elseprintf("-1\n");} return 0; }