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 = 100000 + 10;
int n;
vector<int> G[maxn];
bool del[maxn];
int legs[maxn];
void dfs(int u, int fa) {
  if (G[u].size() <= 2) {
    del[u] = 1;
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (v == fa) continue;
      dfs(v, u);
    }
  }
}
int main() {
  scanf("%d", &n);
  int a, b;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 1; i <= n; i++)
    if (G[i].size() == 1) dfs(i, -1);
  for (int i = 1; i <= n; i++)
    if (!del[i])
      for (int j = 0; j < G[i].size(); j++)
        if (del[G[i][j]]) legs[i] = min(legs[i] + 1, 2);
  for (int i = 1; i <= n; i++)
    if (!del[i]) {
      int cnt = 0;
      for (int j = 0; j < G[i].size(); j++)
        if (!del[G[i][j]] && G[G[i][j]].size() - legs[G[i][j]] > 1) cnt++;
      if (cnt > 2) {
        puts("No");
        return 0;
      }
    }
  puts("Yes");
  return 0;
}
