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 = 1e5 + 5;
int ds = 0;
vector<int> adj[N];
int lvs[N], sz[N], dep[N];
vector<int> twos[N];
int dfs(int v, int p, int n) {
  sz[v] = 1;
  dep[v] = dep[p] + 1;
  int mxd = 0;
  lvs[v] = (adj[v].size() == 1);
  for (auto it : adj[v]) {
    if (it == p) continue;
    int d = dfs(it, v, n);
    lvs[v] += lvs[it];
    if (lvs[it] > 2 or (lvs[it] == 2 and (int) adj[it].size() < 3))
      twos[v].push_back(it);
    sz[v] += sz[it];
    mxd = max(mxd, d);
  }
  if (ds - lvs[v] > 2 or (ds - lvs[v] == 2 and (int) adj[p].size() < 3)) {
    twos[v].push_back(p);
  }
  return mxd + 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b), adj[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if ((int)adj[i].size() == 1) ds++;
  }
  dfs(1, 0, n);
  for (int i = 1; i <= n; i++) {
    if ((int)twos[i].size() > 2) {
      cout << "No\n";
      exit(0);
    }
  }
  cout << "Yes\n";
  return 0;
}
