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;
const int max_n = 100 * 1000 + 10;
vector<int> v[max_n];
bool check = false;
int a[max_n];
int c = 0;
void dfs(int u, int p) {
  int cnt = 0;
  bool t = false;
  for (auto i : v[u])
    if (i != p) {
      dfs(i, u);
      if (a[i] == 1 || v[i].size() > 3) cnt++;
      if (v[i].size() > 2) t = true;
    }
  if (p >= 0) {
    if (cnt > 1) {
      if (c == 0) {
        c++;
        dfs(u, -1);
      } else
        check = true;
    }
    if (cnt > 0) a[u] = 1;
    if (t) a[u] = 1;
  } else {
    if (cnt > 2) check = true;
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  int id = 0;
  for (int i = 0; i < n; i++)
    if (v[i].size() > v[id].size()) id = i;
  if (v[id].size() < 3) {
    cout << "YES" << endl;
    return 0;
  }
  dfs(id, -1);
  if (check) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
}
