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;
int vis[100005];
vector<set<int> > g(100005);
int d[100005];
int f = 0;
int kk = 0;
int dfs(int v, int p) {
  if (kk == 1) return 0;
  if (d[v] > 2) {
    f++;
    kk = 1;
    return 0;
  }
  for (set<int>::iterator it = g[v].begin(); it != g[v].end(); it++) {
    if (*it != p) {
      dfs(*it, v);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int x, y;
  for (int i = 0; i < n + 1; i++) d[i] = 0;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    x--;
    y--;
    d[x]++;
    d[y]++;
    g[x].insert(y);
    g[y].insert(x);
  }
  int flag = 0;
  for (int i = 0; i < n; i++) {
    if (d[i] < 3)
      continue;
    else {
      vector<pair<int, int> > v;
      for (set<int>::iterator it = g[i].begin(); it != g[i].end(); it++) {
        v.push_back({d[*it], *it});
      }
      int fff = 0;
      for (int j = 0; j < v.size(); j++) {
        int u = v[j].second;
        if (d[u] > 3) {
          fff++;
          continue;
        }
        f = 0;
        for (set<int>::iterator it = g[u].begin(); it != g[u].end(); it++) {
          if (*it != i) {
            kk = 0;
            dfs(*it, u);
          }
        }
        if (f != 0) fff++;
      }
      if (fff > 2) {
        flag = 1;
      }
      if (flag == 1) break;
    }
  }
  if (flag == 1)
    cout << "No\n";
  else
    cout << "Yes\n";
  return 0;
}
