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 maxn = 1e5 + 100;
int n, p, q, x[maxn];
set<int> s[maxn];
bool mark[maxn], size[maxn], check = 1;
void dfs(int u) {
  if (size[u] or mark[u] or x[u] > 2) return;
  if (x[u] <= 1)
    x[*s[u].begin()]++;
  else
    mark[*s[u].begin()] = 1;
  s[*s[u].begin()].erase(u);
  if (s[*s[u].begin()].size() == 1) dfs(*s[u].begin());
  size[u] = 1;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int fi, se;
    cin >> fi >> se;
    s[fi].insert(se);
    s[se].insert(fi);
  }
  for (int i = 1; i <= n; i++)
    if (s[i].size() == 1) dfs(i);
  for (int i = 1; i <= n; i++)
    if (!size[i] && s[i].size() > 2) check = 0;
  if (check == 1)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}
