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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int maxn = 1e5 + 10;
long gcd(long a, long b) {
  while (a && b)
    if (a >= b)
      a %= b;
    else
      b %= a;
  return a | b;
}
char used[maxn] = {};
int legs[maxn] = {};
vector<vector<int> > graph;
void dfs(int v, int p = -1) {
  if (int((graph[v]).size()) <= 2) {
    used[v] = 1;
    for (int i : graph[v])
      if (i != p) dfs(i, v);
  }
}
int main() {
  iostream::sync_with_stdio(0);
  int n;
  cin >> n;
  graph.resize(n);
  for (int(i) = 0; (i) < (n - 1); (i)++) {
    int from, to;
    cin >> from >> to;
    from--;
    to--;
    graph[from].push_back((to));
    graph[to].push_back((from));
  }
  for (int(i) = 0; (i) < (n); (i)++)
    if (int((graph[i]).size()) == 1) dfs(i);
  for (int(i) = 0; (i) < (n); (i)++)
    for (int j : graph[i])
      if (used[j]) legs[i] = min(2, legs[i] + 1);
  for (int(i) = 0; (i) < (n); (i)++)
    if (!used[i]) {
      int cnt = 0;
      for (int j : graph[i])
        if (!used[j] && int((graph[j]).size()) - legs[j] > 1) cnt++;
      if (cnt > 2) {
        cout << "NO";
        return 0;
      }
    }
  cout << "YES";
  return 0;
}
