[树的度数] Christmas Spruce

Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex u is called a child of vertex v and vertex v is called a parent of vertex u if there exists a directed edge from v to u. A vertex is called a leaf if it doesn't have children and has a parent.

Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.

The definition of a rooted tree can be found here.

Input

The first line contains one integer n — the number of vertices in the tree (3 ≤ n ≤ 1 000). Each of the next n - 1 lines contains one integer p_i (1 ≤ i ≤ n - 1) — the index of the parent of the i + 1-th vertex (1 ≤ p_i ≤ i).

Vertex 1 is the root. It's guaranteed that the root has at least 2 children.

Output

Print "Yes" if the tree is a spruce and "No" otherwise.

Examples

Input

4
1
1
1

Output

Yes

Input

7
1
1
1
2
2
2

Output

No

Input

8
1
1
1
1
3
3
3

Output

Yes

Note

The first example:

The second example:

It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.

The third example:

题意:问一棵树的所有非叶子结点是否至少有三个叶子
思路:BFS搜索每个非叶子结点的叶子结点的个数,碰到少于3的输出No,否则若都大于等于3个就输出Yes

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
typedef long long ll;
const int amn=1e5+5;
int n,ans=0,m[amn],deep[amn],idx[amn],cnt;
vector<int> eg[amn];
queue<int> q;
int bfs(int rt){
    while(q.size())q.pop();q.push(rt);
    memset(deep,0,sizeof deep);
    memset(idx,0,sizeof idx);
    deep[rt]=1;
    cnt=0;
    while(q.size()){
        int u=q.front();q.pop();
        idx[u]=1;
        cnt=0;
        for(int i=0;i<eg[u].size();i++){
            int v=eg[u][i];
            if(idx[v]||v==u)continue;
            if(!eg[v].size())cnt++; ///统计叶子结点个数
        }
        if(cnt>=3){
                for(int i=0;i<eg[u].size();i++){
                int v=eg[u][i];
                if(idx[v]||v==u)continue;
                if(eg[v].size())q.push(v);
            }
        }
        else return 0;
    }
    return 1;
}
int main(){
    scanf("%d",&n);
    for(int i=2;i<=n;i++){
        scanf("%d",&m[i]);
        eg[m[i]].push_back(i);
    }
    ans=bfs(1);
    if(ans)
    printf("Yes
");
    else printf("No
");
}
/**
题意:问一棵树的所有非叶子结点是否至少有三个叶子
思路:BFS搜索每个非叶子结点的叶子结点的个数,碰到少于3的输出No,否则若都大于等于3个就输出Yes
**/