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 EPS = 1e-10;
const double PI = acos(-1.0);
vector<int> chd[100010];
bool deleted[100010];
int legs[100010];
void dfs(int v, int p) {
  if (int((chd[v]).size()) <= 2) {
    deleted[v] = true;
    for (int to : chd[v])
      if (to != p) {
        legs[to]++;
        dfs(to, v);
      }
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int N;
  cin >> N;
  for (int i = (0); i < (N - 1); ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    chd[a].push_back(b);
    chd[b].push_back(a);
  }
  for (int i = (0); i < (N); ++i)
    if (int((chd[i]).size()) == 1) dfs(i, -1);
  bool ok = true;
  for (int i = (0); i < (N); ++i) {
    if (deleted[i]) continue;
    int cnt = 0;
    for (int to : chd[i])
      if (!deleted[to] && int((chd[to]).size()) - min(legs[to], 2) > 1) ++cnt;
    if (cnt > 2) ok = false;
  }
  cout << (ok ? "Yes" : "No") << endl;
  return 0;
}
