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 maxn = 100 * 1000 + 10;
bool mark[maxn];
int legs[maxn];
vector<int> vr[maxn];
void dfs(int v, int p = 0) {
  if (vr[v].size() <= 2) {
    mark[v] = true;
    for (int i = 0; i < vr[v].size(); i++)
      if (vr[v][i] != p) dfs(vr[v][i], v);
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u;
    vr[v].push_back(u);
    vr[u].push_back(v);
  }
  for (int i = 1; i <= n; i++)
    if (vr[i].size() == 1) dfs(i);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < vr[i].size(); j++)
      if (mark[vr[i][j]] == true) legs[i] = min(legs[i] + 1, 2);
  for (int i = 1; i <= n; i++)
    if (mark[i] == false) {
      int cnt = 0;
      for (int j = 0; j < vr[i].size(); j++)
        if (mark[vr[i][j]] == false &&
            vr[vr[i][j]].size() - legs[vr[i][j]] - 1 > 0)
          cnt++;
      if (cnt > 2) {
        cout << "No\n";
        return 0;
      }
    }
  cout << "Yes\n";
  return 0;
}
