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<int> v[100005];
int n, de[100005], C[100005];
bool mark[100005], ans = true;
void DFS(int now) {
  mark[now] = true;
  for (int i = 0; i < v[now].size(); i++) {
    int to = v[now][i];
    if (mark[to] == true) continue;
    if (v[to].size() > 2) continue;
    DFS(to);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1, a, b; i < n; i++) {
    scanf("%d %d", &a, &b);
    de[a]++, de[b]++;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (int i = 1; i <= n; i++)
    if (de[i] == 1 && mark[i] == false) DFS(i);
  for (int i = 1; i <= n; i++) {
    if (mark[i] == true) continue;
    for (int j = 0; j < v[i].size(); j++) {
      int to = v[i][j];
      if (mark[to] == false) continue;
      C[i] = min(C[i] + 1, 2);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (mark[i] == true) continue;
    int c = 0;
    for (int j = 0; j < v[i].size(); j++) {
      int to = v[i][j];
      if (mark[to] == true) continue;
      if (de[to] > C[to] + 1) c++;
    }
    if (c > 2) ans = false;
    if (ans == false) break;
  }
  if (ans == true)
    puts("Yes");
  else
    puts("No");
  return 0;
}
