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 = 1e5 + 10;
vector<int> G[N];
int T[N], st[N];
queue<int> q;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 0; i < n; ++i) {
    st[i] = G[i].size();
    if (st[i] == 1) q.push(i);
  }
  for (int i = 0; i < n; ++i) T[i] = -1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (G[u].size() == 1) {
      T[u] = 0;
    }
    int cnt[4] = {0, 0, 0, 0};
    for (int i = 0; i < G[u].size(); ++i) {
      int v = G[u][i];
      if (T[v] >= 0) {
        cnt[T[v]]++;
      }
      st[v]--;
      if (st[v] == 1 && T[v] < 0) {
        q.push(v);
      }
    }
    if ((cnt[3] == 1 && cnt[2] > 0) || cnt[3] > 1 || cnt[2] > 2) {
      cout << "No";
      return 0;
    }
    if (cnt[3] == 1 || cnt[2] == 2) {
      T[u] = 3;
    } else if (cnt[2] == 1 || cnt[1] > 0 || cnt[0] > 2) {
      T[u] = 2;
    } else if (cnt[0] == 2) {
      T[u] = 1;
    } else {
      T[u] = 0;
    }
  }
  cout << "Yes";
}
