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> v[100201];
bool deleted[1000201];
int t[1000201];
int pr[1000201];
void dfs(int u) {
  if ((int)v[u].size() <= 2) {
    deleted[u] = true;
    for (int j = 0; j < (int)v[u].size(); j++) {
      int h = v[u][j];
      if (!deleted[h]) {
        pr[h] = u;
        dfs(h);
      }
    }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  pr[0] = -1;
  for (int i = 0; i < n; i++) {
    if ((int)v[i].size() == 1) {
      dfs(i);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < (int)v[i].size(); j++) {
      int h = v[i][j];
      if (deleted[h]) {
        t[i] = min(t[i] + 1, 2);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (!deleted[i]) {
      int count = 0;
      for (int j = 0; j < (int)v[i].size(); j++) {
        int h = v[i][j];
        if (!deleted[h]) {
          if ((int)v[h].size() - t[h] > 1) {
            count++;
          }
        }
      }
      if (count > 2) {
        cout << "No" << endl;
        return 0;
      }
    }
  }
  cout << "Yes" << endl;
  return 0;
}
