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<vector<int> > graph;
vector<int> marked, legs;
void dfs(int v, int p) {
  if (graph[v].size() <= 2) {
    marked[v] = true;
    for (int x : graph[v]) {
      if (x == p) {
        continue;
      }
      dfs(x, v);
    }
  }
}
void solve() {
  int n, f, t;
  cin >> n;
  marked.resize(n + 1, 0);
  legs.resize(n + 1, 0);
  graph.resize(n + 1);
  for (int i = 0; i < n - 1; ++i) {
    cin >> f >> t;
    graph[f].push_back(t);
    graph[t].push_back(f);
  }
  for (int i = 1; i <= n; ++i) {
    if (graph[i].size() == 1) {
      dfs(i, -1);
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int x : graph[i]) {
      if (marked[x]) {
        legs[i] = min(2, legs[i] + 1);
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (!marked[i]) {
      int cnt = 0;
      for (int x : graph[i]) {
        if (!marked[x] && graph[x].size() - legs[x] > 1) {
          ++cnt;
        }
      }
      if (cnt > 2) {
        cout << "No";
        return;
      }
    }
  }
  cout << "Yes";
}
int main() {
  solve();
  return 0;
}
