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 N = 100010;
int inf = 1e9;
int mod = 1e9 + 7;
vector<vector<int>> graph;
vector<bool> hot;
void dfs(int cur, int par = -1) {
  if ((int)graph[cur].size() <= 2) {
    hot[cur] = true;
    for (int u : graph[cur]) {
      if (u != par) {
        dfs(u, cur);
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  graph.resize(n + 1);
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  hot.assign(n + 1, false);
  for (int i = 1; i <= n; i++) {
    if ((int)graph[i].size() == 1) {
      dfs(i);
    }
  }
  vector<int> cnt_branch(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    for (int u : graph[i]) {
      if (hot[u]) {
        cnt_branch[i]++;
      }
    }
    cnt_branch[i] = min(cnt_branch[i], 2);
  }
  for (int i = 1; i <= n; i++) {
    if (!hot[i]) {
      int temp = 0;
      for (int u : graph[i]) {
        if (!hot[u] && (int)graph[u].size() - cnt_branch[u] > 1) {
          temp++;
        }
      }
      if (temp > 2) {
        cout << "No\n";
        return 0;
      }
    }
  }
  cout << "Yes\n";
  return 0;
}
