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 maxn = 1e5 + 5;
vector<int> g[maxn];
bool mark[maxn];
int legs[maxn];
void dfs(int x, int y) {
  if (g[x].size() <= 2) {
    mark[x] = true;
    for (int to : g[x]) {
      if (to != y) {
        dfs(to, x);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int _n(n - 1), i(0); i < _n; i++) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int _n(n), v(0); v < _n; v++)
    if (!mark[v] && g[v].size() == 1) dfs(v, -1);
  for (int _n(n), v(0); v < _n; v++) {
    for (int u : g[v]) {
      if (mark[u]) {
        legs[v] = min(2, legs[v] + 1);
      }
    }
  }
  for (int _n(n), v(0); v < _n; v++)
    if (!mark[v]) {
      int cnt = 0;
      for (int u : g[v]) {
        if (!mark[u] && (int)g[u].size() - legs[u] > 1) {
          ++cnt;
        }
      }
      if (cnt > 2) {
        cout << "No\n";
        return 0;
      }
    }
  cout << "Yes\n";
  return 0;
}
