Limak is a little bear who learns to draw. People usually start with houses, fences and flowers but why would bears do it? Limak lives in the forest and he decides to draw a tree.

Recall that tree is a connected graph consisting of n vertices and n - 1 edges.

Limak chose a tree with n vertices. He has infinite strip of paper with two parallel rows of dots. Little bear wants to assign vertices of a tree to some n distinct dots on a paper so that edges would intersect only at their endpoints — drawn tree must be planar. Below you can see one of correct drawings for the first sample test.

<image>

Is it possible for Limak to draw chosen tree?

Input

The first line contains single integer n (1 ≤ n ≤ 105).

Next n - 1 lines contain description of a tree. i-th of them contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) denoting an edge between vertices ai and bi. It's guaranteed that given description forms a tree.

Output

Print "Yes" (without the quotes) if Limak can draw chosen tree. Otherwise, print "No" (without the quotes).

Examples

Input

8
1 2
1 3
1 6
6 4
6 7
6 5
7 8


Output

Yes


Input

13
1 2
1 3
1 4
2 5
2 6
2 7
3 8
3 9
3 10
4 11
4 12
4 13


Output

No
#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> graph[100009];
bool vis[100009];
int leg[100009];
void dfs(int i, int from) {
  if (graph[i].size() > 2) return;
  vis[i] = 1;
  for (int f = 0; f < graph[i].size(); f++)
    if (graph[i][f] != from) dfs(graph[i][f], i);
}
int main() {
  cin >> n;
  for (int f = 1; f < n; f++) {
    int a, b;
    scanf("%d %d", &a, &b);
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  for (int f = 1; f <= n; f++)
    if (graph[f].size() == 1) dfs(f, f);
  for (int f = 1; f <= n; f++)
    for (int f1 = 0; f1 < graph[f].size(); f1++) leg[f] += vis[graph[f][f1]];
  bool flag = 0;
  for (int f = 1; f <= n; f++) {
    int temp = 0;
    if (vis[f]) continue;
    for (int f1 = 0; f1 < graph[f].size(); f1++) {
      if (vis[graph[f][f1]]) continue;
      if (graph[graph[f][f1]].size() - min(2, leg[graph[f][f1]]) > 1) temp++;
    }
    if (temp > 2) {
      cout << "No\n";
      return 0;
    }
  }
  cout << "Yes\n";
}
