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;
template <class T>
T inline sqr(T x) {
  return x * x;
}
const long double pi = 3.1415926535897932384626433832795;
const long double eps = 1e-8;
const int N = 100500;
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<set<int>> S(n);
  for (int i = 0; i < (int)(n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    S[u].insert(v);
    S[v].insert(u);
  }
  auto st = S;
  for (int v = 0; v < (int)(n); ++v)
    if (S[v].size() == 2) {
      vector<int> to((S[v]).begin(), (S[v]).end());
      int x = to[0], y = to[1];
      S[x].erase(v);
      S[y].erase(v);
      S[v].clear();
      S[x].insert(y);
      S[y].insert(x);
    }
  for (int v = 0; v < (int)(n); ++v) {
    int fat = 0;
    for (int u : S[v]) {
      if (S[u].size() == 1) continue;
      if (S[u].size() == 3) {
        vector<pair<int, int>> to;
        for (auto z : S[u]) to.push_back(make_pair(S[z].size(), z));
        sort((to).begin(), (to).end());
        if (to[0].first == 1 && to[1].first == 1 && st[u].count(v)) {
          continue;
        }
      }
      fat++;
    }
    if (fat > 2) {
      puts("No");
      return 0;
    }
  }
  puts("Yes");
  return 0;
}
