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;
vector<int> e[100005];
int del[100005], leg[100005];
void dfs(int u, int fa) {
  if (e[u].size() <= 2)
    del[u] = 1;
  else
    return;
  for (auto v : e[u]) {
    if (v == fa) continue;
    dfs(v, u);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  while (cin >> n) {
    int x, y;
    for (int i = 1; i <= n; i++) e[i].clear();
    memset(del, 0, sizeof(del));
    memset(leg, 0, sizeof(leg));
    for (int i = 1; i <= n - 1; i++) {
      cin >> x >> y;
      e[x].push_back(y);
      e[y].push_back(x);
    }
    for (int i = 1; i <= n; i++)
      if (e[i].size() == 1) dfs(i, -1);
    for (int i = 1; i <= n; i++) {
      for (auto j : e[i]) {
        if (del[j]) leg[i] = min(leg[i] + 1, 2);
      }
    }
    for (int i = 1; i <= n; i++) {
      if (del[i]) continue;
      int cnt = 0;
      for (auto v : e[i])
        if (!del[v] && e[v].size() - leg[v] > 1) cnt++;
      if (cnt > 2) {
        cout << "No" << endl;
        return 0;
      }
    }
    cout << "Yes" << endl;
  }
  return 0;
}
