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;
set<int> v[100003];
int a[100003], b[100003], c[100003], n;
void dfs(int x) {
  if (b[x] || c[x] || a[x] > 2) return;
  if (a[x] < 2)
    a[*v[x].begin()]++;
  else
    b[*v[x].begin()] = 1;
  v[*v[x].begin()].erase(x);
  if (v[*v[x].begin()].size() == 1) dfs(*v[x].begin());
  c[x] = 1;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    v[a].insert(b);
    v[b].insert(a);
  }
  for (int i = 0; i < n; i++)
    if (v[i].size() == 1) dfs(i);
  for (int i = 0; i < n; i++)
    if (c[i] == 0 && v[i].size() > 2) {
      cout << "No";
      return 0;
    }
  cout << "Yes";
}
