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 NMAX = 1e5 + 10;
int N;
bool removed[NMAX], _removed[NMAX];
vector<int> T[NMAX], edges;
int main() {
  ios_base::sync_with_stdio(false);
  int a, b, i;
  cin >> N;
  for (i = 1; i < N; ++i) {
    cin >> a >> b;
    T[a].emplace_back(b);
    T[b].emplace_back(a);
  }
  for (i = 1; i <= N; ++i) {
    if ((int(T[i].size())) == 1) {
      removed[i] = 1;
      int now = T[i][0], last = i;
      while ((int(T[now].size())) <= 2) {
        if ((int(T[now].size())) == 1) {
          cout << "Yes\n";
          return 0;
        }
        removed[now] = 1;
        if (T[now][0] == last) {
          last = now;
          now = T[now][1];
        } else if (T[now][1] == last) {
          last = now;
          now = T[now][0];
        }
      }
    }
  }
  for (i = 1; i <= N; ++i) {
    edges.clear();
    for (const int &j : T[i])
      if (!removed[j]) edges.emplace_back(j);
    swap(T[i], edges);
    if ((int(T[i].size())) == 1 && (int(edges.size())) <= 3) _removed[i] = 1;
  }
  for (i = 1; i <= N; ++i) {
    edges.clear();
    for (const int &j : T[i])
      if (removed[j] == 0 && _removed[j] == 0) edges.emplace_back(j);
    swap(T[i], edges);
    if ((int(T[i].size())) > 2) {
      cout << "No\n";
      return 0;
    }
  }
  cout << "Yes\n";
  return 0;
}
