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;
vector<vector<int> > node;
bool del[100005];
int leg[100005];
void dfs(int u, int pre) {
  if (node[u].size() <= 2) {
    del[u] = 1;
    for (int i = 0; i < node[u].size(); i++) {
      if (node[u][i] == pre) continue;
      dfs(node[u][i], u);
    }
  }
}
int main() {
  int n, u, v;
  scanf("%d", &n);
  node.clear();
  node.resize(n + 1);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    node[u].push_back(v);
    node[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    if (node[i].size() == 1) dfs(i, 0);
  }
  for (int i = 1; i <= n; i++) {
    if (del[i]) continue;
    for (int j = 0; j < node[i].size(); j++) {
      int k = node[i][j];
      if (del[k]) leg[i]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (del[i]) continue;
    int cnt = 0;
    for (int j = 0; j < node[i].size(); j++) {
      int k = node[i][j];
      if (!del[k] && node[k].size() > min(leg[k], 2) + 1) cnt++;
    }
    if (cnt > 2) {
      printf("No\n");
      return 0;
    }
  }
  printf("Yes\n");
  return 0;
}
