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 <class T, class U>
inline void chmin(T &t, U f) {
  if (t > f) t = f;
}
template <class T, class U>
inline void chmax(T &t, U f) {
  if (t < f) t = f;
}
const int SIZE = 100000;
int N;
vector<int> G[SIZE];
int deg[SIZE];
bool used[SIZE];
void dfs(int v, int p) {
  if (G[v].size() >= 3) {
    deg[v]--;
    return;
  }
  used[v] = true;
  for (int to : G[v]) {
    if (to == p) continue;
    dfs(to, v);
  }
}
signed main() {
  scanf("%d", &N);
  for (int i = 0; i < (N - 1); i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 0; i < (N); i++) deg[i] = G[i].size();
  for (int i = 0; i < (N); i++)
    if (G[i].size() == 1) dfs(i, -1);
  for (int i = 0; i < (N); i++)
    if (!used[i]) {
      if (G[i].size() >= 3) {
        for (int to : G[i]) {
          if (G[to].size() == 3 && deg[to] == 1) {
            used[to] = true;
            deg[i]--;
          }
        }
      }
    }
  bool ok = true;
  for (int i = 0; i < (N); i++)
    if (!used[i]) ok &= deg[i] <= 2;
  if (ok)
    puts("Yes");
  else
    puts("No");
  return 0;
}
