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;
using namespace std;
int n, p, q, x[100005], y[100005], z[100005], o;
set<int> v[100005];
inline void d(int a) {
  if (z[a] || y[a] || x[a] > 2) return;
  if (x[a] < 2)
    x[*v[a].begin()]++;
  else
    y[*v[a].begin()] = 1;
  v[*v[a].begin()].erase(a);
  if (v[*v[a].begin()].size() == 1) d(*v[a].begin());
  z[a] = 1;
}
int main() {
  cin >> n;
  for (int a = 1; a <= n - 1; a++) {
    cin >> p >> q;
    v[p].insert(q);
    v[q].insert(p);
  }
  for (int a = 1; a <= n; a++)
    if (v[a].size() == 1) d(a);
  for (int a = 1; a <= n; a++)
    if (!z[a] && v[a].size() > 2) o = 1;
  if (o)
    cout << "No";
  else
    cout << "Yes";
}
