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;
void DFS(int vertex, const vector<unordered_set<int>> &graph,
         vector<char> &used) {
  used[vertex] = 2;
  for (int to : graph[vertex]) {
    if (!used[to] && graph[to].size() < 3) {
      DFS(to, graph, used);
    }
  }
}
int main() {
  size_t n;
  cin >> n;
  vector<unordered_set<int>> graph(n);
  for (size_t i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    graph[x].insert(y);
    graph[y].insert(x);
  }
  vector<size_t> size(n);
  vector<char> used(n);
  for (size_t i = 0; i < n; ++i) {
    size[i] = graph[i].size();
    if (graph[i].size() == 1) {
      DFS(i, graph, used);
    }
  }
  for (int vertex = 0; vertex < n; ++vertex) {
    if (used[vertex] == 2) {
      continue;
    }
    int counter = 0;
    for (int to : graph[vertex]) {
      if (used[to] == 2) {
        ++counter;
      }
    }
    if (counter + 1 == graph[vertex].size() && graph[vertex].size() <= 3) {
      used[vertex] = 1;
    }
  }
  for (size_t vertex = 0; vertex < n; ++vertex) {
    if (used[vertex] == 0) {
      int counter = 0;
      for (int to : graph[vertex]) {
        if (used[to] == 0) {
          ++counter;
        }
      }
      if (counter > 2) {
        cout << "No\n";
        return 0;
      }
    }
  }
  cout << "Yes\n";
  return 0;
}
