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;
vector<set<long long int>> G;
vector<set<long long int>> oriG;
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  long long int n;
  cin >> n;
  G.resize(n);
  for (long long int i = 0; i < n - 1; ++i) {
    long long int a, b;
    cin >> a >> b;
    --a;
    --b;
    G[a].insert(b);
    G[b].insert(a);
  }
  oriG = G;
  if (n == 1) {
    cout << "Yes\n";
    return 0;
  }
  for (long long int v = 0; v < n; ++v) {
    if (G[v].size() == 2) {
      auto it = G[v].begin();
      long long int a = *it;
      ++it;
      long long int b = *it;
      G[v].clear();
      G[a].erase(v);
      G[b].erase(v);
      G[a].insert(b);
      G[b].insert(a);
    }
  }
  for (long long int v = 0; v < n; ++v) {
    long long int cnt = 0;
    for (long long int x : G[v]) {
      if (G[x].size() >= 2) {
        if (G[x].size() == 3 && oriG[v].count(x)) {
          bool ok = true;
          for (long long int y : G[x]) {
            if (y == v) continue;
            if (G[y].size() != 1) ok = false;
          }
          if (ok) continue;
        }
        ++cnt;
      }
    }
    if (cnt >= 3) {
      cout << "No\n";
      return 0;
    }
  }
  cout << "Yes\n";
  return 0;
}
