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>
std::set<int> node[131072];
std::vector<int> sta;
int col[131072];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    node[u].insert(v);
    node[v].insert(u);
    ++col[u], ++col[v];
  }
  for (int i = 1; i <= n; ++i) {
    if (col[i] > 1) continue;
    int u = i;
    while (col[u] < 3) {
      if (node[u].empty()) break;
      int v = *node[u].begin();
      node[u].clear();
      node[v].erase(u);
      u = v;
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (col[i] == 3 && node[i].size() == 1) sta.push_back(i);
  }
  while (!sta.empty()) {
    int u = sta.back(), v = *node[u].begin();
    sta.pop_back();
    node[u].clear();
    node[v].erase(u);
  }
  int u = 1;
  while (node[u].size() != 1 && u <= n) ++u;
  if (u != n + 1) {
    int k;
    while (node[u].size() == 1) {
      k = u, u = *node[u].begin();
      node[u].erase(k);
    }
  }
  puts(node[u].empty() ? "Yes" : "No");
  return 0;
}
