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 A, class B>
ostream& operator<<(ostream& out, const pair<A, B>& a) {
  return out << "(" << a.first << ", " << a.second << ")";
}
template <int>
ostream& operator<<(ostream& os, const vector<int>& v) {
  os << "[";
  for (int i = 0; i < v.size(); ++i) {
    if (v[i] != 1000000001)
      os << v[i];
    else
      os << "INF";
    if (i != v.size() - 1) os << ", ";
  }
  os << "]\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (int i = 0; i < v.size(); ++i) {
    os << v[i];
    ;
    if (i != v.size() - 1) os << ", ";
  }
  os << "]\n";
  return os;
}
const int N = 1e5 + 10;
vector<int> adj[N] = {};
vector<int> path(N);
vector<int> good(N);
vector<int> par(N);
int root = -1;
void dfs(int u = 0, int p = -1) {
  par[u] = p;
  bool isLeaf = true;
  int paths = 0;
  for (auto v : adj[u]) {
    if (v - p) {
      isLeaf = false;
      dfs(v, u);
      if (adj[u].size() == 2 && path[v] && u != root) {
        path[u] = 1;
        good[u] = 1;
        paths++;
      } else if (path[v]) {
        paths++;
      }
    }
  }
  if (isLeaf) {
    path[u] = 1;
    good[u] = 1;
  }
  if (paths <= 2 && adj[u].size() == paths + 1) {
    good[u] = 1;
  }
}
int main() {
  {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  };
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 0; i < n; i++) {
    if (adj[i].size() >= 3) {
      root = i;
      break;
    }
  }
  if (root == -1) {
    cout << "Yes" << endl;
    return 0;
  }
  dfs(root);
  bool boo = true;
  for (int i = 0; i < n; i++) {
    if (good[i]) continue;
    int notGood = 0;
    for (auto v : adj[i]) {
      if (!good[v]) {
        notGood++;
      }
    }
    if (notGood > 2) boo = false;
  }
  if (boo)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}
