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 M = 1e5 + 5;
int deg[M];
vector<int> edge[M];
bool mark[M];
int del[M];
void dfs(int x, int f) {
  mark[x] = true;
  for (int i = 0; i < edge[x].size(); ++i) {
    if (edge[x][i] == f)
      continue;
    else if (deg[edge[x][i]] <= 2)
      dfs(edge[x][i], x);
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    deg[x]++;
    deg[y]++;
    edge[x].push_back(y);
    edge[y].push_back(x);
  }
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) dfs(i, -1);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < edge[i].size(); ++j) {
      if (mark[edge[i][j]]) del[i] = min(del[i] + 1, 2);
    }
  }
  bool ok = true;
  for (int i = 1; i <= n; ++i) {
    int cnt = 0;
    for (int j = 0; j < edge[i].size(); ++j) {
      if (deg[edge[i][j]] - del[edge[i][j]] > 1) cnt++;
    }
    if (cnt > 2) ok = false;
  }
  if (ok)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}
