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 mxn = 100006;
int n;
set<int> e[mxn];
int f[mxn];
int f3[mxn];
bool v[mxn];
void EXEC() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int L, R;
    scanf("%d%d", &L, &R);
    e[L].insert(R);
    e[R].insert(L);
  }
  for (int i = 1; i <= n; ++i) {
    f[i] = (e[i].size() >= 3);
    f3[i] = (e[i].size() == 3);
  }
  deque<int> que;
  for (int i = 1; i <= n; ++i)
    if (!f[i] && e[i].size() <= 1) que.push_back(i);
  while (!que.empty()) {
    int x = que.front();
    que.pop_front();
    v[x] = true;
    if (!e[x].empty()) {
      int y = *e[x].begin();
      e[x].erase(y);
      e[y].erase(x);
      if (!f[y] && e[y].size() <= 1) que.push_back(y);
    }
  }
  for (int i = 1; i <= n; ++i)
    if (!v[i] && f3[i] && e[i].size() <= 1) que.push_back(i);
  for (auto x : que) {
    v[x] = true;
    if (!e[x].empty()) {
      int y = *e[x].begin();
      e[x].erase(y);
      e[y].erase(x);
    }
  }
  if ([] {
        for (int i = 1; i <= n; ++i)
          if (!v[i])
            ({
              if (!(e[i].size() <= 2)) return false;
            });
        return true;
      }())
    cout << "Yes\n";
  else
    cout << "No\n";
}
int main() {
  EXEC();
  return 0;
}
