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<int> a[100001];
vector<bool> del(100001, false);
vector<int> legs(100001, 0);
void dfs(int node, int parent = 0) {
  if ((int)((a[node]).size()) < 3) {
    del[node] = true;
    for (int b : a[node]) {
      if (b != parent) dfs(b, node);
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (__typeof(n) i = (1); i != (n); i++) {
    int p, q;
    cin >> p >> q;
    a[p].push_back(q);
    a[q].push_back(p);
  }
  for (__typeof(n + 1) i = (1); i != (n + 1); i++) {
    if ((int)((a[i]).size()) == 1) {
      dfs(i);
    }
  }
  for (__typeof(n + 1) i = (1); i != (n + 1); i++) {
    for (int b : a[i]) {
      if (del[b]) legs[i] = min(legs[i] + 1, 2);
    }
  }
  for (__typeof(n + 1) i = (1) - ((1) > (n + 1));
       i != (n + 1) - ((1) > (n + 1)); i += 1 - 2 * ((1) > (n + 1))) {
    if (!del[i]) {
      int cnt = 0;
      for (int b : a[i]) {
        if (!del[b] && (int)((a[b]).size()) - legs[b] > 1) cnt++;
      }
      if (cnt > 2) {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  cout << "YES" << endl;
  return 0;
}
