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;
int cc[100010];
int hasmul[100010];
string res = "Yes";
vector<int> adj[100010];
bool dfs(int idx, int par) {
  hasmul[idx] = (adj[idx].size() > 3) ? 1 : 0;
  for (int i = 0; i < adj[idx].size(); i++) {
    int v = adj[idx][i];
    if (v != par) {
      dfs(v, idx);
      if (hasmul[v] == 1) cc[idx]++;
      if (hasmul[v]) hasmul[idx] = 1;
    }
  }
  if (hasmul[idx] == 0 && adj[idx].size() == 3) hasmul[idx] = 2;
  return hasmul[idx];
}
void dfs2(int idx, int par, int mul) {
  for (int i = 0; i < adj[idx].size(); i++) {
    int v = adj[idx][i];
    if (v != par) {
      int tmp = (mul > 0 ? 1 : 0);
      if (cc[idx] > hasmul[v] || adj[idx].size() > 3) tmp = 1;
      if (tmp == 0 && adj[idx].size() == 3) tmp = 2;
      dfs2(v, idx, tmp);
    }
  }
  if (mul == 2) mul = 0;
  if (cc[idx] + mul > 2) res = "No";
}
int main() {
  cin >> n;
  int a, b;
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(0, -1);
  for (int i = 0; i < n; i++)
    if (hasmul[i] == 2) hasmul[i] = 0;
  dfs2(0, -1, 0);
  cout << res << endl;
  return 0;
}
