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;
mt19937 rng(423978);
const int mn = 5e5 + 10;
set<int> g[mn];
int ori[mn];
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    g[a].insert(b);
    g[b].insert(a);
  }
  stack<int> s;
  for (int i = 1; i <= n; i++) {
    ori[i] = g[i].size();
    if (ori[i] == 1) s.push(i);
  }
  while (s.size()) {
    int x = s.top();
    s.pop();
    int y = *g[x].begin();
    g[x].erase(y);
    g[y].erase(x);
    if (g[y].size() == 1 && ori[y] <= 2) s.push(y);
  }
  for (int i = 1; i <= n; i++)
    if (g[i].size() == 1 && ori[i] == 3) {
      int j = *g[i].begin();
      g[i].erase(j);
      g[j].erase(i);
    }
  for (int i = 1; i <= n; i++)
    if (g[i].size() >= 3) {
      printf("No");
      return 0;
    }
  printf("Yes");
}
