Codeforces Round #398 (Div. 2) C. Garland

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：给你一棵树，树上n个点，每个点都有权值，问能否把某两个点上的边去了，使树分成三等分，输出那两个点。

思路：题目本身并不难想，一看就知道是树形dp，不过坑略多，同一棵子树的情况要考虑清楚，还有其他乱七八糟的，还是看代码吧。

#include<cstdio>  
#include<algorithm>  
#include<cstring>  
#include<iostream>  
#include<cmath>  
#include<queue>  
#include<functional>  
typedef long long LL;
using namespace std;
#define maxn 1000005
#define ll l,mid,now<<1  
#define rr mid+1,r,now<<1|1  
#define lson l1,mid,l2,r2,now<<1  
#define rson mid+1,r1,l2,r2,now<<1|1  
#define inf 0x3f3f3f3f  
const int mod = 1e9 + 7;
int value[maxn], head[maxn], mark[maxn];
int ans[5], sum, len;
struct node{
	int v, next;
}p[maxn];
void dfs(int x, int fa){
	for (int i = head[x]; ~i; i = p[i].next){
		dfs(p[i].v, x);
		mark[x] = max(mark[x], mark[p[i].v]);
		value[x] += value[p[i].v];
	}
	if (len == 2||!fa||!x)
		return;
	if ((!mark[x] && value[x] == sum / 3) ||(mark[x]&& value[x] == sum / 3 * 2)){
		ans[len++] = x;   
		mark[x] = 1;
		if (len == 2)
			return;
	}
}
int main(){
	int n;
	memset(head, -1, sizeof(head));
	scanf("%d", &n);
	for (int i = 1; i <= n; i++){
		int u;
		scanf("%d%d", &u,&value[i]);
		sum += value[i];
		p[i].v = i;
		p[i].next = head[u];
		head[u] = i;
	}
	if (sum % 3)
		printf("-1\n");
	else{
		dfs(0, -1);
		if (len == 2)
			printf("%d %d\n", ans[0], ans[1]);
		else
			printf("-1\n");
	}
}