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 = 1e9;
const long long int INFLL = 1e18;
const double EPS = 1e-8;
template <class T>
T &chmin(T &a, const T &b) {
  return a = min(a, b);
}
template <class T>
T &chmax(T &a, const T &b) {
  return a = max(a, b);
}
long long int in() {
  long long int a;
  cin >> a;
  return a;
}
const int MAX_N = 100010;
vector<long long int> G[MAX_N];
bool ans = true;
int a[MAX_N];
int dfs(int x, int p) {
  for (auto w : G[x])
    if (w != p) a[x] += dfs(w, x);
  return a[x] + (G[x].size() >= 3 ? 1 : 0);
}
void dfs2(int x, int p, bool t) {
  int res = 0;
  if (p != -1)
    if (G[p].size() >= 4 || t) res++;
  for (auto w : G[x])
    if (w != p)
      if (G[w].size() >= 4 || a[w] >= 1) res++;
  if (res >= 3) ans = false;
  for (auto w : G[x])
    if (w != p)
      dfs2(w, x,
           t || (p != -1 && G[p].size() >= 3) ||
               a[x] - (a[w] + (G[w].size() >= 3 ? 1 : 0)) >= 1);
}
int main() {
  int n = in();
  for (int i = (0); i < (int)(n - 1); i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(0, -1);
  dfs2(0, -1, false);
  if (ans)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}
