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 = 1e5 + 5;
int n, root;
vector<int> adj[N];
bool ok;
int dfs(int v, int p = -1) {
  int ret = 0;
  int cnt = 2;
  if (v == root) cnt++;
  for (int u : adj[v]) {
    if (u == p) continue;
    int res = dfs(u, v);
    if (res) ret = 2;
    if (res == 2) cnt--;
  }
  if (cnt <= 0) {
    ok = false;
  }
  if (adj[v].size() == 3) ret = max(1, ret);
  if (adj[v].size() > 3) ret = 2;
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  ok = true;
  root = 0;
  for (int i = 0; i < n; i++)
    if (adj[root].size() < adj[i].size()) root = i;
  dfs(root);
  if (!ok && adj[root].size() == 3) {
    ok = true;
    for (int u : adj[root]) {
      if (adj[u].size() == 3) {
        root = u;
        break;
      }
    }
    dfs(root);
  }
  if (ok)
    puts("Yes");
  else
    puts("No");
  return 0;
}
