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;
template <typename T>
inline T abs(T t) {
  return t < 0 ? -t : t;
}
const unsigned long long modn = 1000000007;
inline unsigned long long mod(unsigned long long x) { return x % modn; }
const int MAX = 100009;
int deg[MAX], ok[MAX];
vector<int> adj[MAX];
void dfs(int u, int p) {
  if (deg[u] > 3) return;
  ok[u]++;
  if (deg[u] == 3) return;
  for (int v : adj[u])
    if (v != p) dfs(v, u);
}
int main() {
  int i, n, a, b;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%d %d", &a, &b);
    adj[--a].push_back(--b);
    adj[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  for (i = 0; i < n; i++)
    if (deg[i] == 1) dfs(i, -1);
  for (i = 0; i < n; i++)
    if (deg[i] == 3 && ok[i] <= 1) ok[i] = 0;
  for (i = 0; i < n; i++) {
    if (ok[i]) continue;
    int ct = 0;
    for (int v : adj[i])
      if (!ok[v]) ct++;
    if (ct > 2) break;
  }
  if (i < n)
    puts("No");
  else
    puts("Yes");
}
