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 + 10;
vector<int> a[maxn];
int d[maxn], n;
int vis[maxn];
void dfs(int x, int f) {
  for (int i = 0; i < a[x].size(); i++) {
    int u = a[x][i];
    if (u == f) continue;
    if (d[u] <= 2) {
      vis[u] = 1;
      dfs(u, x);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
    d[x]++;
    d[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (d[i] == 1) vis[i] = 1, dfs(i, 0);
  for (int i = 1; i <= n; i++) {
    int cnt = 0;
    if (d[i] != 3) continue;
    for (int j = 0; j < a[i].size(); j++)
      if (vis[a[i][j]]) cnt++;
    if (cnt == 2) vis[i] = 1;
  }
  for (int i = 1; i <= n; i++) {
    int cnt = 0;
    for (int j = 0; j < a[i].size(); j++)
      if (vis[a[i][j]]) cnt++;
    if (d[i] - cnt > 2) {
      puts("No");
      return 0;
    }
  }
  puts("Yes");
  return 0;
}
