Codeforces Round #530 (Div. 1) 1098A Sum in the tree

A. Sum in the tree

Mitya has a rooted tree with nn vertices indexed from 11 to nn, where the root has index 11. Each vertex vv initially had an integer number av≥0av≥0 written on it. For every vertex vv Mitya has computed svsv: the sum of all values written on the vertices on the path from vertex vv to the root, as well as hvhv — the depth of vertex vv, which denotes the number of vertices on the path from vertex vv to the root. Clearly, s1=a1s1=a1 and h1=1h1=1.

Then Mitya erased all numbers avav, and by accident he also erased all values svsv for vertices with even depth (vertices with even hvhv). Your task is to restore the values avav for every vertex, or determine that Mitya made a mistake. In case there are multiple ways to restore the values, you’re required to find one which minimizes the total sum of values avav for all vertices in the tree.

Input

The first line contains one integer nn — the number of vertices in the tree (2≤n≤1052≤n≤105). The following line contains integers p2p2, p3p3, … pnpn, where pipi stands for the parent of vertex with index ii in the tree (1≤pi<i1≤pi<i). The last line contains integer values s1s1, s2s2, …, snsn (−1≤sv≤109−1≤sv≤109), where erased values are replaced by −1−1.

Output

Output one integer — the minimum total sum of all values avav in the original tree, or −1−1 if such tree does not exist.

Examples

input

5

1 1 1 1

1 -1 -1 -1 -1

output

1

input

5

1 2 3 1

1 -1 2 -1 -1

output

2

input

3

1 2

2 -1 1

output

-1

这道题时说有一颗树，树有N个节点，每个节点有一个值Vi，沿着根节点到I节点求和得到SUMi，前缀和，现在删去vi，且删去了偶数深度节点的sumi，vi>=0，sumi<=1e9.现是否存在一棵这样的树使得各点vi的和相加最小，不存在就是，存在子节点sumi小于父节点sumi使得子节点vi为负值。当为偶数节点时，他要被做差分，他也要与父节点做差分，如果他有子节点时，他们几个构成贪心的机制，代价=∑（sum[子节点]-sum[当前节点]）+sum[当前节点]-sum[父节点] 可知当子节点大于1的时候sum[当前]越大越好，最大的符合条件的sum[当前]是 sum【子节点】的最小值，当没有子节点时，显而易见等于父节点的sum【父节点】时代价最小，当只有一个子节点时，代价去sum【子】与sum【父】带价相同，归到第一类，可以写代码了。

    #include<bits/stdc++.h>
    #define Swap(a,b) a^=b^=a^=b
    #define cini(n) scanf("%d",&n)
    #define cinl(n) scanf("%lld",&n)
    #define cinc(n) scanf("%c",&n)
    #define speed ios_base::sync_with_stdio(0); // 切不可用scnaf；
    #define Max(a,b) a>b?a:b
    #define Min(a,b) a<b?a:b
    using namespace std;
    typedef long long ll;
    const int INF=0x3f3f3f3f;
    const int maxn=1e5+10;
    int fa[maxn];
    ll sum[maxn];
    int main()
    {
        speed
        int n;
        ll ans=0;
        cin>>n;
        for(int i=2;i<=n;i++)
        {
            cin>>fa[i];
        }
        for(int i=1;i<=n;i++)
        {
            cin>>sum[i];
            if(sum[i]==-1) sum[i]=1e9+1;
           // sum[i]=sum[i]==-1?1e9+1:sum[i];
        }
       // for(int i=1;i<=n;i++) cout<<sum[i]<<' '<<endl;
       // cout<<endl;
        for(int i=2;i<=n;i++)
        {
            sum[fa[i]]=Min(sum[fa[i]],sum[i]);
        }
        //    for(int i=1;i<=n;i++) cout<<sum[i]<<' '<<endl;
        for(int i=2;i<=n;i++)
        {
            int temp=sum[i]-sum[fa[i]];
            if(temp<0) return cout<<-1<<endl,0;
            if(sum[i]<=1e9) ans+=temp;
        }
        cout<<sum[1]+ans<<endl;
    }