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;
const int N = 100000 + 10;
int n;
int father[N];
vector<int> adj[N];
int bfs(int s) {
  vector<int> que;
  father[s] = -1;
  que.push_back(s);
  int ret = -1;
  for (int i = 0; i < que.size(); ++i) {
    int u = que[i];
    if (adj[u].size() >= 3) {
      ret = u;
    }
    for (int e = 0; e < adj[u].size(); ++e) {
      int v = adj[u][e];
      if (v == father[u]) continue;
      father[v] = u;
      que.push_back(v);
    }
  }
  return ret;
}
int on[N];
void solve() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u - 1].push_back(v - 1);
    adj[v - 1].push_back((u - 1));
  }
  int flag = false;
  for (int i = 0; i < n; ++i) {
    if (adj[i].size() >= 3) {
      flag = true;
    }
  }
  if (!flag) {
    puts("Yes");
    return;
  }
  int s = bfs(0);
  int t = bfs(s);
  for (int u = t; u != -1; u = father[u]) {
    on[u] = true;
  }
  int cnt = 0;
  for (int u = t; u != -1; u = father[u]) {
    if (adj[u].size() >= 3) ++cnt;
    for (int e = 0; e < adj[u].size(); ++e) {
      int v = adj[u][e];
      if (on[v] || adj[v].size() > 3) continue;
      cnt += adj[v].size() == 3;
    }
  }
  int total = 0;
  for (int i = 0; i < n; ++i) {
    if (adj[i].size() >= 3) ++total;
  }
  if (cnt != total) {
    puts("No");
    return;
  }
  puts("Yes");
}
int main() {
  solve();
  return 0;
}
