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;
vector<vector<int> > g;
vector<int> u, rem;
vector<set<int> > mem;
void go(int i, int par = -1) {
  if (g[i].size() > 2) return;
  rem[i] = 1;
  for (int j = 0; j < g[i].size(); j++) {
    if (g[i][j] != par) go(g[i][j], i);
  }
}
int main() {
  int n;
  while (cin >> n) {
    g.clear();
    g.resize(n);
    for (int i = 0; i < n - 1; i++) {
      int x, y;
      scanf("%d %d", &x, &y);
      ;
      g[x - 1].push_back(y - 1);
      g[y - 1].push_back(x - 1);
    }
    u.clear();
    u.resize(n, 0);
    rem = u;
    mem.clear();
    mem.resize(n);
    for (int i = 0; i < n; i++) {
      if (g[i].size() == 1) go(i);
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < g[i].size(); j++) u[i] += rem[g[i][j]];
    }
    bool ok = 1;
    for (int i = 0; i < n && ok; i++) {
      if (!rem[i]) {
        int tmp = 0;
        for (int j = 0; j < g[i].size(); j++) {
          if (g[g[i][j]].size() - min(u[g[i][j]], 2) > 1) tmp++;
        }
        if (tmp > 2) ok = 0;
      }
    }
    ok ? cout << "Yes\n" : cout << "No\n";
  }
  return 0;
}
