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;
queue<int> todo;
vector<int> edges[100002];
bool considered[100002];
int fromHere[100002];
int dfs(int where, int ignore, bool subtract) {
  int order;
  if (subtract) {
    order = edges[where].size() - 1;
  } else {
    order = edges[where].size();
  }
  for (int i = 0; i < edges[where].size(); i++) {
    if (edges[where][i] != ignore) {
      if (fromHere[edges[where][i]] == 0)
        fromHere[edges[where][i]] = dfs(edges[where][i], where, false);
      order = max(fromHere[edges[where][i]], order);
    }
  }
  return order;
}
int main() {
  int n;
  int startHere = -1;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    edges[a].push_back(b);
    edges[b].push_back(a);
    if (edges[a].size() >= 3) startHere = a;
    if (edges[b].size() >= 3) startHere = b;
  }
  if (startHere == -1) {
    cout << "Yes" << endl;
    return 0;
  }
  todo.push(startHere);
  while (!todo.empty()) {
    int a = 0;
    startHere = todo.front();
    considered[startHere] = true;
    todo.pop();
    for (int i = 0; i < edges[startHere].size(); i++) {
      if (dfs(edges[startHere][i], startHere, true) > 2) {
        a++;
        if (!considered[edges[startHere][i]]) todo.push(edges[startHere][i]);
      }
    }
    if (a > 2) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  return 0;
}
