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 MAXN = 1e5 + 5;
vector<int> G[MAXN];
int n, leg[MAXN];
bool mark[MAXN];
void init() {
  cin >> n;
  for (int u, v, i = 1; i < n; i++) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
}
void dfs(int u, int f) {
  if (G[u].size() <= 2) {
    mark[u] = 1;
    for (int i = 0; i < G[u].size(); i++)
      if (G[u][i] != f) dfs(G[u][i], u);
  }
  return;
}
void solve() {
  for (int i = 1; i <= n; i++)
    if (G[i].size() == 1) dfs(i, 0);
  for (int u = 1; u <= n; u++)
    if (!mark[u])
      for (int i = 0; i < G[u].size(); i++)
        if (mark[G[u][i]]) leg[u]++;
  for (int u = 1; u <= n; u++)
    if (!mark[u]) {
      int cnt = 0;
      for (int i = 0; i < G[u].size(); i++)
        if (!mark[G[u][i]])
          if (G[G[u][i]].size() > min(leg[G[u][i]], 2) + 1) cnt++;
      if (cnt > 2) {
        cout << "No";
        return;
      }
    }
  cout << "Yes" << endl;
}
int main() {
  init();
  solve();
  return 0;
}
