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;
long long gcd(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int maxn = 100010;
int n;
vector<int> adj[maxn];
int vis[maxn];
int cnt[maxn];
void dfs(int u, int p) {
  vis[u] = 1;
  for (int i = (0); i < (int((adj[u]).size())); i++) {
    int v = adj[u][i];
    if (v != p && int((adj[v]).size()) == 2) dfs(v, u);
  }
}
void solve() {
  memset(vis, 0, sizeof(vis));
  memset(cnt, 0, sizeof(cnt));
  scanf("%d", &n);
  for (int i = (0); i < (n - 1); i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = (0); i < (n); i++)
    if (int((adj[i]).size()) == 1) dfs(i, -1);
  for (int i = (0); i < (n); i++)
    for (int j = (0); j < (int((adj[i]).size())); j++) {
      int k = adj[i][j];
      if (vis[k]) cnt[i]++;
    }
  for (int i = (0); i < (n); i++) {
    int tmp = 0;
    for (int j = (0); j < (int((adj[i]).size())); j++) {
      int k = adj[i][j];
      if (int((adj[k]).size()) - min(2, cnt[k]) > 1) tmp++;
    }
    if (tmp > 2) {
      printf("No");
      return;
    }
  }
  printf("Yes");
}
int main() {
  solve();
  return 0;
}
