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;
long long MOD = 1000000007;
vector<vector<long long> > g;
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long N;
  cin >> N;
  g.resize(N);
  long long res = 0;
  long long a, b;
  for (long long i = 0; i < N - 1; i++) {
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  queue<long long> qu;
  vector<long long> cnt(N, 0);
  vector<long long> V(N, 0);
  for (long long i = 0; i < N; i++) {
    if (g[i].size() == 1) {
      qu.push(i);
      cnt[i] = 1;
      V[i] = 1;
    }
  }
  while (qu.size()) {
    a = qu.front();
    qu.pop();
    if (cnt[a] == 1 && (long long)g[a].size() <= 2) {
      V[a] = 1;
      for (auto i : g[a]) {
        if (V[i] == 0) {
          cnt[i]++;
          qu.push(i);
        }
      }
    }
  }
  for (long long i = 0; i < N; i++) {
    if (V[i] == 0) {
      if ((cnt[i] == 1 && g[i].size() == 2) ||
          (cnt[i] == 2 && g[i].size() == 3)) {
        V[i] = 1;
      }
    }
  }
  bool ok = true;
  long long c1 = 0;
  long long vs = 0;
  long long es = 0;
  for (long long i = 0; i < N; i++) {
    if (V[i] == 0) {
      vs++;
      long long z = 0;
      for (auto j : g[i]) {
        if (V[j] == 0) {
          z++;
        }
      }
      es += z;
      if (z > 2) {
        ok = false;
      }
    }
  }
  if (vs > 0 && es / 2 != vs - 1) {
    ok = false;
  }
  if (ok) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
}
