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;
int hcf(int a, int b) {
  if (!a)
    return (b);
  else
    return (hcf(b % a, a));
}
int lcm(int a, int b) { return ((a * b) / hcf(a, b)); }
vector<vector<long long>> adj;
set<long long> done;
vector<long long> endpoints;
vector<long long> status;
void dfs(long long node) {
  if ((long long)(adj[node].size()) > 2) {
    return;
  }
  if (done.count(node)) return;
  done.insert(node);
  if ((long long)(adj[node].size()) == 1) {
    status[node] = 0;
    dfs(adj[node][0]);
    return;
  }
  if ((long long)(adj[node].size()) == 2) {
    status[node] = 1;
    dfs(adj[node][0]);
    dfs(adj[node][1]);
  }
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n;
  cin >> n;
  adj = vector<vector<long long>>(n, vector<long long>(0, -1));
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  endpoints = {};
  for (long long i = 0; i < n; i++)
    if ((long long)(adj[i].size()) == 1) endpoints.push_back(i);
  status = vector<long long>(n, -1);
  for (long long endpoint : endpoints) {
    done = {};
    dfs(endpoint);
  }
  for (long long i = 0; i < n; i++) {
    if ((long long)(adj[i].size()) == 3) {
      long long cntLin = 0;
      for (long long node : adj[i]) {
        cntLin += (status[node] == 0 || status[node] == 1);
      }
      if (cntLin == 3) {
        cout << ("Yes") << "\n";
        return 0;
      }
      if (cntLin == 2) {
        status[i] = 2;
        continue;
      }
    }
  }
  unordered_set<long long> internal_path = {};
  for (long long i = 0; i < n; i++)
    if (status[i] == -1) internal_path.insert(i);
  for (long long internal_node : internal_path) {
    long long cnt_internal_connections = 0;
    for (long long ngh : adj[internal_node]) {
      if (status[ngh] == -1) cnt_internal_connections++;
    }
    if (cnt_internal_connections > 2) {
      cout << ("No") << "\n";
      return 0;
    }
  }
  cout << ("Yes") << "\n";
  return 0;
  return 0;
}
