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;
int main() {
  int n, m;
  cin >> n;
  vector<vector<int> > G(n);
  vector<int> P(n, 0);
  vector<vector<int> > k(n, vector<int>(5, 0));
  priority_queue<pair<int, int> > Q;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    G[x].push_back(y);
    G[y].push_back(x);
    ++P[x];
    ++P[y];
  }
  for (int i = 0; i < n; ++i) {
    if (P[i] == 1) {
      Q.push(pair<int, int>(-1, i));
    }
  }
  vector<int> T = P;
  bool xec = true;
  while (!Q.empty()) {
    int x = Q.top().second;
    Q.pop();
    if (k[x][1] > 0 or k[x][2] > 2) {
      xec = false;
      k[x][0] = 1;
    } else if (k[x][2] == 1)
      k[x][0] = 2;
    else if (k[x][2] == 2)
      k[x][0] = 1;
    else if (k[x][3] > 0 or k[x][4] > 2)
      k[x][0] = 2;
    else if (k[x][4] == 2)
      k[x][0] = 3;
    else
      k[x][0] = 4;
    int count = 0;
    for (int i = 0; i < G[x].size(); ++i) {
      int y = G[x][i];
      if (P[y] > 1) {
        --P[y];
        k[y][k[x][0]]++;
        if (P[y] == 1) Q.push(pair<int, int>(-T[y], y));
        ++count;
      }
    }
    if (count == 0 and k[x][0] == 1) {
      for (int i = 0; i < G[x].size(); ++i) {
        int y = G[x][i];
        if (k[y][0] == 2) ++count;
      }
      if (count > 2) xec = false;
    } else if (count == 0 and k[x][0] == 2) {
      for (int i = 0; i < G[x].size(); ++i) {
        int y = G[x][i];
        if (k[y][0] == 1) xec = false;
      }
    }
  }
  int count = 0;
  if (count > 1) xec = false;
  if (xec)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}
