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;
template <class T, class L>
bool smax(T &x, L y) {
  return x < y ? (x = y, 1) : 0;
}
template <class T, class L>
bool smin(T &x, L y) {
  return y < x ? (x = y, 1) : 0;
}
const int maxn = 1e5 + 17;
int n, my[maxn];
vector<int> g[maxn];
bool block[maxn];
void go_up(int v, int p = -1) {
  if (g[v].size() > 2) return;
  block[v] = 1;
  for (auto u : g[v])
    if (u != p) go_up(u, v);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int v, u, i = 1; i < n; i++)
    cin >> v >> u, g[--v].push_back(--u), g[u].push_back(v);
  for (int i = 0; i < n; i++)
    if (g[i].size() == 1) go_up(i, -1);
  for (int v = 0; v < n; v++)
    for (auto u : g[v])
      if (block[u]) my[v]++;
  for (int v = 0; v < n; v++)
    if (!block[v]) {
      int cnt = 0;
      for (auto u : g[v]) cnt += !block[u] && g[u].size() - min(my[u], 2) > 1;
      if (cnt > 2) return cout << "No\n", 0;
    }
  cout << "Yes\n";
  return 0;
}
