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 N = 100000;
vector<int> ladj[N];
bool vis[N];
int arr[N];
void dfs(int u, int p) {
  if ((int)ladj[u].size() > 2) return;
  vis[u] = true;
  for (int i = 0; i < (int)ladj[u].size(); i++) {
    int v = ladj[u][i];
    if (v != p) dfs(v, u);
  }
}
bool solve() {
  int n;
  if (scanf("%d", &n) == EOF) return false;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    ladj[--x].push_back(--y);
    ladj[y].push_back(x);
  }
  for (int i = 0; i < n; i++)
    if ((int)ladj[i].size() == 1) dfs(i, -1);
  for (int i = 0; i < n; i++)
    for (int k = 0; k < (int)ladj[i].size(); k++) {
      int j = ladj[i][k];
      if (vis[j] && arr[i] < 2) arr[i]++;
    }
  for (int i = 0; i < n; i++)
    if (!vis[i]) {
      int x = 0;
      for (int k = 0; k < (int)ladj[i].size(); k++) {
        int j = ladj[i][k];
        if (!vis[j] && (int)ladj[j].size() - arr[j] > 1) x++;
      }
      if (x > 2) {
        puts("No");
        return true;
      }
    }
  puts("Yes");
  return true;
}
int main() {
  while (solve())
    ;
  return 0;
}
