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::vector<int> G[100033];
bool del[100033];
int legs[100033];
void dfs(int a, int par = 0) {
  if (G[a].size() <= 2) {
    del[a] = 1;
    for (int i = (0), __ = (G[a].size()); i < __; ++i)
      if (G[a][i] != par) dfs(G[a][i], a);
  }
}
int n;
int a, b;
int cnt;
int main() {
  scanf("%d", &n);
  for (int i = (0), __ = (n - 1); i < __; ++i) {
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int a = (1), __ = (n); a <= __; ++a)
    if (G[a].size() == 1) dfs(a);
  for (int a = (1), __ = (n); a <= __; ++a)
    for (int j = (0), __ = (G[a].size()); j < __; ++j)
      if (del[G[a][j]]) {
        legs[a] = std::min(legs[a] + 1, 2);
      }
  for (int a = (1), __ = (n); a <= __; ++a)
    if (!del[a]) {
      cnt = 0;
      for (int j = (0), __ = (G[a].size()); j < __; ++j)
        if (!del[G[a][j]] && G[G[a][j]].size() - legs[G[a][j]] > 1) {
          ++cnt;
        }
      if (cnt > 2) {
        puts("No");
        return 0;
      }
    }
  puts("Yes");
  return 0;
}
