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 = 1e5;
int N, A[MAX_N + 1], L[MAX_N + 1];
bool D[MAX_N + 1];
vector<int> E[MAX_N + 1];
void dfs(int n, int p = 0) {
  int i;
  if (E[n].size() <= 2) {
    D[n] = true;
    for (i = 0; i < E[n].size(); i++) {
      if (E[n][i] != p) dfs(E[n][i], n);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int i, j, a, b, c;
  cin >> N;
  for (i = 1; i < N; i++) {
    cin >> a >> b;
    E[a].push_back(b);
    E[b].push_back(a);
  }
  for (i = 1; i <= N; i++)
    if (E[i].size() == 1) {
      dfs(i);
    }
  for (i = 1; i <= N; i++)
    for (j = 0; j < E[i].size(); j++)
      if (D[E[i][j]]) {
        L[i] = min(L[i] + 1, 2);
      }
  for (i = 1; i <= N; i++)
    if (!D[i]) {
      c = 0;
      for (j = 0; j < E[i].size(); j++)
        if (!D[E[i][j]] && E[E[i][j]].size() - L[E[i][j]] > 1) {
          c++;
        }
      if (c > 2) {
        cout << "NO\n";
        return 0;
      }
    }
  cout << "YES\n";
  return 0;
}
