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 limite = 1000000;
int n;
vector<int> g[limite];
int eliminado[limite];
int hoja[limite];
int abajo[limite];
int arriba[limite];
int arhoja[limite];
int ararriba[limite];
void marcahoja(int u) {
  if (hoja[u]) return;
  vector<int> &ar = g[u];
  if (int(ar.size()) > 2) return;
  hoja[u] = 1;
  if (int(ar.size()) == 0) return;
  marcahoja(ar[0]);
  if (int(ar.size()) <= 1) return;
  marcahoja(ar[1]);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int u = 1; u <= n; u++)
    if (int(g[u].size()) <= 1) marcahoja(u);
  for (int u = 1; u <= n; u++) {
    vector<int> &ar = g[u];
    for (int i = 0; i < int(ar.size()); i++) {
      int v = ar[i];
      arhoja[u] += hoja[v];
    }
    abajo[u] =
        hoja[u] or (arhoja[u] <= 2 and int(g[u].size()) - arhoja[u] <= 1);
    arriba[u] = 1 - abajo[u];
  }
  for (int u = 1; u <= n; u++) {
    if (arriba[u]) {
      vector<int> &ar = g[u];
      for (int i = 0; i < int(ar.size()); i++) {
        int v = ar[i];
        ararriba[u] += arriba[v];
      }
      if (ararriba[u] >= 3) {
        cout << "No" << endl;
        exit(0);
      }
    }
  }
  cout << "Yes" << endl;
}
