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 N;
vector<int> G[100001];
int P[100001];
int calc(int n, int p) {
  int& r = P[n];
  int chInP = 0;
  int childRoot = 0;
  for (int ch : G[n]) {
    if (ch == p) continue;
    calc(ch, n);
    if (P[ch] == -1) return r = -1;
    if (P[ch] != 0) chInP++;
    if (P[ch] == 1 || P[ch] == 3) childRoot = P[ch];
  }
  if (chInP >= 3) return r = -1;
  if (childRoot && (r != 0 || chInP >= 2)) return r = -1;
  if (childRoot == 1 && (G[n].size() >= 4)) return r = -1;
  if (childRoot == 3 && (G[n].size() >= 3)) return r = -1;
  if (childRoot) return r = 3;
  if (chInP >= 2) return r = 1;
  if (chInP >= 1) return r = 2;
  if (p != 0 && G[n].size() >= 4) return r = 2;
  if (p != 0 && r == 0 && G[n].size() >= 3) P[p] = 2;
  return r;
}
int main() {
  istream& in = cin;
  in >> N;
  for (int i = 1; i < N; i++) {
    int a, b;
    in >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
    P[i] = 0;
  }
  int r = calc(1, 0);
  cout << ((r != -1) ? "Yes\n" : "No\n");
}
