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;
vector<int> E[101010];
int have3[101010];
int dfs(int cur, int pre) {
  for (auto& tar : E[cur])
    if (tar != pre) have3[cur] += dfs(tar, cur);
  return have3[cur] += (E[cur].size() >= 3);
}
void dfs2(int cur, int pre, int n3) {
  int ng = (n3 >= 2 || (n3 == 1 && pre != -1 &&
                        (E[pre].size() == 2 || E[pre].size() == 4)));
  for (auto& tar : E[cur])
    if (tar != pre)
      if (have3[tar] >= 2 ||
          (have3[tar] == 1 && (E[tar].size() == 2 || E[tar].size() == 4)))
        ng++;
  if (ng >= 3) (void)printf("No\n"), exit(0);
  for (auto& tar : E[cur])
    if (tar != pre) dfs2(tar, cur, have3[cur] + n3 - have3[tar]);
}
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N;
  for (i = 0; i < N - 1; i++) {
    cin >> x >> y;
    E[x - 1].push_back(y - 1);
    E[y - 1].push_back(x - 1);
  }
  dfs(0, -1);
  dfs2(0, -1, 0);
  (void)printf("Yes\n");
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false);
  for (i = 0; i < argc - 1; i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < s.size(); i++) ungetc(s[s.size() - 1 - i], stdin);
  solve();
  return 0;
}
