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 INF = std::numeric_limits<int>::max() / 3;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
vector<vector<int> > tree;
vector<int> del;
vector<int> leg;
void dfs(int u, int p = 0) {
  del[u] = 1;
  for (int i = 0; i < tree[u].size(); ++i) {
    if (tree[u][i] != p && tree[tree[u][i]].size() <= 2) dfs(tree[u][i], u);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  tree.assign(n + 1, vector<int>());
  del.assign(n + 1, 0);
  leg.assign(n + 1, 0);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    tree[a].push_back(b);
    tree[b].push_back(a);
  }
  for (int i = 1; i <= n; ++i) {
    if (!del[i] && tree[i].size() == 1) dfs(i);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < tree[i].size(); ++j) {
      int v = tree[i][j];
      if (del[v]) leg[i] = min(leg[i] + 1, 2);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (!del[i]) {
      int cnt = 0;
      for (int j = 0; j < tree[i].size(); ++j) {
        int v = tree[i][j];
        if (!del[v]) {
          if (tree[v].size() - leg[v] > 1) cnt++;
        }
      }
      if (cnt > 2) {
        cout << "No" << endl;
        return 0;
      }
    }
  }
  cout << "Yes" << endl;
  return 0;
}
