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;
inline string get_string() {
  char ch[1000000];
  scanf("%s", ch);
  return ch;
}
vector<vector<int>> graph(100010);
bool del[100010];
int legs[100010];
void dfs(int vertex, int err = -1) {
  if (graph[vertex].size() <= 2) {
    del[vertex] = true;
    for (int i = 0; i < graph[vertex].size(); i++)
      if (err != graph[vertex][i]) dfs(graph[vertex][i], vertex);
  }
}
void no() {
  cout << "No"
       << "\n";
  exit(0);
}
void yes() {
  cout << "Yes"
       << "\n";
  exit(0);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i + 1 < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (graph[i].size() == 1) dfs(i);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < graph[i].size(); j++)
      if (del[graph[i][j]]) legs[i] = legs[i] < 2 ? legs[i] + 1 : 2;
  for (int i = 1; i <= n; i++)
    if (!del[i]) {
      int cnt = 0;
      for (int j = 0; j < graph[i].size(); j++)
        if (del[graph[i][j]] == false &&
            graph[graph[i][j]].size() - legs[graph[i][j]] > 1)
          cnt++;
      if (cnt > 2) no();
    }
  yes();
}
