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 double PI = acos(-1);
const long long mod = 1e9 + 7;
const long long N = 1e5 + 5, K = 105;
const int si = (1 << 20);
const long long add = 1e4;
const long long M = 100 + 5;
const double INFdb = 1e10;
int n;
vector<int> v[N];
bool check[N];
int legs[N];
void dfs(int u, int par) {
  if (v[u].size() <= 2) {
    check[u] = 1;
    for (int x : v[u]) {
      if (x != par) {
        dfs(x, u);
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 1; i <= n; ++i) {
    if ((int)v[i].size() == 1) dfs(i, 0);
  }
  for (int i = 1; i <= n; ++i) {
    for (int x : v[i]) {
      if (check[x]) {
        legs[i] = min(legs[i] + 1, 2);
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (!check[i]) {
      int cnt = 0;
      for (int x : v[i]) {
        if (!check[x] && v[x].size() - legs[x] > 1) {
          ++cnt;
        }
      }
      if (cnt > 2) {
        return cout << "NO", 0;
      }
    }
  }
  cout << "YES";
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
