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 N = 100005;
int legs[N];
bool hide[N];
vector<vector<int> > adj;
void dfs(int v, int par = 0) {
  if (adj[v].size() <= 2u) {
    hide[v] = true;
    for (int i : adj[v]) {
      if (i != par) {
        dfs(i, v);
      }
    }
  }
}
int main() {
  int n, a, b;
  scanf("%d", &n);
  adj.resize(n + 1);
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 1; i <= n; ++i) {
    if (adj[i].size() == 1u) {
      dfs(i);
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j : adj[i]) {
      if (hide[j]) {
        legs[i] = min(2, legs[i] + 1);
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (!hide[i]) {
      int count = 0;
      for (int j : adj[i]) {
        if (!hide[j] && (int)adj[j].size() - legs[j] > 1) {
          ++count;
        }
      }
      if (count > 2) {
        printf("No\n");
        return 0;
      }
    }
  }
  printf("Yes\n");
  return 0;
}
