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 long long Maxn = 1e5 + 7;
const long long Max = 4e3 + 7;
const long long Mod = 1e9 + 7;
const long long Inf = 1e9 + 7;
vector<long long> V[Maxn];
long long d[Maxn];
bool mark[Maxn];
void dfs(long long u, long long par) {
  if (V[u].size() > 2) return;
  mark[u]++;
  for (long long i = 0; i < V[u].size(); i++) {
    long long v = V[u][i];
    if (par != v) dfs(v, u);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    V[u].push_back(v);
    V[v].push_back(u);
  }
  for (long long u = 1; u <= n; u++)
    if (V[u].size() == 1) dfs(u, -1);
  for (long long u = 1; u <= n; u++)
    for (long long i = 0; i < V[u].size(); i++) {
      long long v = V[u][i];
      if (mark[v]) d[u] = min(d[u] + 1, 2ll);
    }
  for (long long u = 1; u <= n; u++)
    if (!mark[u]) {
      long long ted = 0;
      for (long long i = 0; i < V[u].size(); i++) {
        long long v = V[u][i];
        if (!mark[v] && V[v].size() - d[v] > 1) ted++;
      }
      if (ted > 2) {
        cout << "No" << endl;
        return 0;
      }
    }
  cout << "Yes" << endl;
  return 0;
}
