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 main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  long long int N;
  cin >> N;
  vector<vector<long long int> > nexts(N);
  for (long long int n = 0; n < N - 1; ++n) {
    long long int a, b;
    cin >> a >> b;
    --a;
    --b;
    nexts[a].push_back(b);
    nexts[b].push_back(a);
  }
  vector<long long int> todo;
  set<long long int> good;
  for (long long int n = 0; n < N; ++n)
    if (nexts[n].size() == 1) todo.push_back(n);
  while (todo.size()) {
    long long int n = todo.back();
    todo.pop_back();
    if (nexts[n].size() <= 2) {
      good.insert(n);
      for (long long int next : nexts[n]) {
        if (good.find(next) == good.end()) todo.push_back(next);
      }
    }
  }
  vector<long long int> half_good;
  for (long long int n = 0; n < N; ++n) {
    if (nexts[n].size() == 3) {
      long long int ngood = 0;
      for (long long int next : nexts[n])
        if (good.find(next) != good.end()) ++ngood;
      if (ngood >= 2) half_good.push_back(n);
    }
  }
  for (long long int h : half_good) good.insert(h);
  bool ok = true;
  for (long long int n = 0; n < N; ++n) {
    long long int nbad = 0;
    for (long long int next : nexts[n])
      if (good.find(next) == good.end()) ++nbad;
    if (nbad >= 3) {
      ok = false;
      break;
    }
  }
  cout << (ok ? "Yes" : "No") << endl;
  return 0;
}
