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 long long int mod = 1e9 + 7;
vector<set<long long int>> adj;
long long int del[100005] = {0}, leg[100005] = {0};
void dfs(long long int src, long long int par) {
  if (adj[src].size() > 2) {
    leg[src]++;
    return;
  }
  del[src] = 1;
  for (auto i : adj[src]) {
    if (i != par) dfs(i, src);
  }
}
int main() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(25);
  long long int n, x, y;
  cin >> n;
  adj.resize(n);
  for (int i = 0; i < 100005; ++i) del[i] = leg[i] = 0;
  for (int i = 1; i < n; ++i) {
    cin >> x >> y;
    --x, --y;
    adj[x].insert(y);
    adj[y].insert(x);
  }
  for (int i = 0; i < n; ++i) {
    if (adj[i].size() == 1) {
      dfs(i, -1);
    }
  }
  bool flag = true;
  for (int i = 0; i < n; ++i) {
    if (del[i]) continue;
    long long int cnt = 0;
    for (auto j : adj[i]) {
      if (del[j]) continue;
      if (adj[j].size() - min(2LL, leg[j]) > 1) cnt++;
    }
    if (cnt > 2) flag = false;
  }
  if (flag)
    cout << "Yes\n";
  else
    cout << "No\n";
  return 0;
}
