As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

For a tree T, let F(T,i) be the distance between vertice 1 and vertice i.(The length of each edge is 1). 

Two trees A and B are similiar if and only if the have same number of vertices and for each i meet F(A,i)=F(B,i). 

Two trees A and B are different if and only if they have different numbers of vertices or there exist an number i which vertice i have different fathers in tree A and tree B when vertice 1 is root.

Tree A is special if and only if there doesn‘t exist an tree B which A and B are different and A and B are similiar.

Now he wants to know if a tree is special.

It is too difficult for Rikka. Can you help her?

There are no more than 100 testcases. 

For each testcase, the first line contains a number n(1≤n≤1000).

Then n?1 lines follow. Each line contains two numbers u,v(1≤u,v≤n) , which means there is an edge between u and v.

For each testcase, if the tree is special print "YES" , otherwise print "NO".

3
1 2
2 3
4
1 2
2 3
1 4

YES
NO

Hint
For the second testcase, this tree is similiar with the given tree:
4
1 2
1 4
3 4
 

BestCoder Round #53 (div.2)

中文题意：

众所周知，萌萌哒六花不擅长数学，所以勇太给了她一些数学问题做练习，其中有一道是这样的：

对于一棵树$T$,令$F(T,i)$为点1到点$i$的最短距离（边长是1）. 

两棵树$A$和$B$是相似的当且仅当他们顶点数相同且对于任意的$i$都有$F(A,i)=F(B,i)$.

两棵树$A$和$B$是不同的当且仅当他们定点数不同或者存在一个$i$使得以1号点为根的时候$i$在两棵树中的父亲不同。

一棵树$A$是特殊的当且仅当不存在一棵和它不同的树和它相似。

现在勇太想知道一棵树到底是不是特殊的。

当然，这个问题对于萌萌哒六花来说实在是太难了，你可以帮帮她吗？

数据组数不超过100组。每组数据的第一行一个整数$n(2\le n\le 1000)$。

接下来$n?1$行。每行两个整数$u,v(1\le u,v\le n)$，代表给定树上的一条边。

对于每一组数据，如果给定树是特殊的输出"YES"否则输出"NO"。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<stdlib.h>

using namespace std;

int n,m;
int v[1010];
int map[1001][1001];

int main() {
    while(scanf("%d",&n)!=EOF) {
        memset(map,0,sizeof(map));
        memset(v,0,sizeof(v));
        int x,y;
        int pf;
        for(int i=0; i<n-1; i++) {
            scanf("%d%d",&x,&y);
            map[x][y] = 1;
            map[y][x] = 1;
            if(x == 1) {
                pf = y;
            } else if(y == 1) {
                pf = x;
            }
            v[x]++;
            v[y]++;
        }
        int rt = 0;
        int cnt = 0,ans = 0,pt = 0;
        for(int i=1; i<=n; i++) {
            if(v[i] == 1) {
                ans++;
            } else if(v[i] == 2) {
                cnt++;
            }else{
                rt = i;
                pt++;
            }
        }
        if(pt>=2){
            printf("NO\n");
            continue;
        }
        if(v[1]!=1) {
            if(v[1] == n-1) {
                printf("YES\n");
            } else {
                printf("NO\n");
            }
        }else{
            if(cnt + ans +1 >= n && v[pf] <=2){
                int flag = 0;
                if(pt>0){
                    for(int i=1;i<=n;i++){
                        if(map[rt][i] == 1 && v[i]!=1){
                            flag++;
                        }
                    }
                }
                if(flag <= 1){
                    printf("YES\n");
                }else{
                    printf("NO\n");
                }

            }else{
                if(v[pf] == n-1){
                    printf("YES\n");
                    continue;
                }
                printf("NO\n");
            }
        }

    }
    return 0;
}