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 oo = 1e9 + 7;
vector<vector<int> > G;
vector<int> leg, del;
int n;
void dfs(int v, int p = -1) {
  if (G[v].size() <= 2) {
    del[v] = 1;
    for (int i = 0; i < G[v].size(); ++i) {
      int u = G[v][i];
      if (u == p) continue;
      dfs(u, v);
    }
  }
}
void LoadGraph() {
  cin >> n;
  G.assign(n + 1, vector<int>());
  ;
  leg.assign(n + 1, 0);
  ;
  del.assign(n + 1, 0);
  ;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i < n + 1; ++i)
    if (G[i].size() == 1) dfs(i);
}
int main() {
  ios_base::sync_with_stdio(false);
  LoadGraph();
  for (int v = 1; v < n + 1; ++v)
    for (int i = 0; i < G[v].size(); ++i) {
      int u = G[v][i];
      if (del[u]) leg[v] = min(leg[v] + 1, 2);
    }
  for (int v = 1; v < n + 1; ++v) {
    int cnt = 0;
    for (int i = 0; i < G[v].size(); ++i) {
      int u = G[v][i];
      if (!del[u] && G[u].size() - leg[u] > 1) ++cnt;
    }
    if (cnt > 2) return cout << "No\n", 0;
  }
  cout << "Yes\n";
}
