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 MXN = 1e5 + 10;
long long n, Root;
long long Val[MXN], d[MXN], Bad[MXN], Sub[3][MXN];
vector<long long> adj[MXN];
void DFS(long long u, long long par) {
  for (auto v : adj[u]) {
    if (v == par) continue;
    DFS(v, u), Bad[u] = max(Bad[u], Bad[v]);
  }
  if (Bad[u] <= 1 && d[u] - (u != Root) <= 1)
    Val[u] = 1;
  else if (Bad[u] <= 1 && d[u] - (u != Root) <= 2)
    Val[u] = 2;
  Bad[u] = max(Bad[u], d[u] - (u != Root));
}
bool dfs(long long u, long long par, long long t) {
  for (auto v : adj[u]) {
    if (v == par) continue;
    Sub[Val[v]][u]++;
  }
  if (Sub[0][u] > t) return 0;
  for (auto v : adj[u]) {
    if (Val[v] || v == par) continue;
    long long nw =
        max(1ll, t - 1 + (Sub[0][u] == 1 && Sub[2][u] == 0 && Sub[1][u] <= 2));
    if (u != Root) nw = 1;
    if (!dfs(v, u, nw)) return 0;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v), adj[v].push_back(u);
    d[u]++, d[v]++;
  }
  for (int i = 1; i <= n; i++) {
    if (d[i] >= 3) {
      Root = i;
      break;
    }
  }
  if (!Root) return cout << "Yes" << '\n', 0;
  DFS(Root, 0);
  cout << (dfs(Root, 0, 2) ? "Yes" : "No") << '\n';
  return 0;
}
