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 = 100010;
vector<int> G[MAXN];
bool del[MAXN];
int leg[MAXN];
void dfs(int v, int p = 0) {
  del[v] = true;
  for (int ch : G[v]) {
    if (ch == p) continue;
    if (G[ch].size() <= 2) dfs(ch, v);
  }
}
int main() {
  int N;
  cin >> N;
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 1; i <= N; i++)
    if (G[i].size() == 1) dfs(i);
  for (int i = 1; i <= N; i++) {
    for (int ch : G[i])
      if (del[ch]) leg[i]++;
  }
  for (int i = 1; i <= N; i++) {
    if (del[i]) continue;
    int cnt = 0;
    for (int ch : G[i])
      if (!del[ch]) {
        if (G[ch].size() - min(2, leg[ch]) > 1) cnt++;
      }
    if (cnt > 2) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  return 0;
}
