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 MaxN = 1e5 + 5;
int n, del[MaxN];
vector<int> G[MaxN];
template <typename T>
void Exit(T msg) {
  cout << msg << endl;
  exit(0);
}
void Init() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    G[a].emplace_back(b);
    G[b].emplace_back(a);
  }
}
void Search(int x, int f = -1) {
  if (G[x].size() <= 2) {
    del[x] = true;
    for (int nxt : G[x])
      if (nxt != f) Search(nxt, x);
  }
}
void Main() {
  for (int i = 1; i <= n; ++i)
    if (G[i].size() == 1) Search(i);
  for (int i = 1; i <= n; ++i)
    if (G[i].size() >= 3) {
      int ndeg = 0;
      for (int nxt : G[i])
        if (!del[nxt]) ++ndeg;
      if (G[i].size() == 3 && ndeg == 1) del[i] = true;
    }
  for (int i = 1; i <= n; ++i)
    if (!del[i]) {
      int ndeg = 0;
      for (int nxt : G[i])
        if (!del[nxt]) ++ndeg;
      if (ndeg > 2) Exit("No");
    }
  Exit("Yes");
}
int main() {
  Init();
  Main();
  return 0;
}
