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 nax = 1e5 + 5;
vector<int> w[nax];
bool del[nax];
int legs[nax];
void dfs(int a, int par = 0) {
  if (w[a].size() <= 2) {
    del[a] = true;
    for (int i = 0; i < w[a].size(); i++) {
      int b = w[a][i];
      if (b != par) dfs(b, a);
    }
  }
}
int main() {
  int n, b;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    w[a].push_back(b);
    w[b].push_back(a);
  }
  for (int a = 1; a <= n; ++a)
    if (w[a].size() == 1) dfs(a);
  for (int a = 1; a <= n; ++a) {
    legs[a] = 0;
    for (int i = 0; i < w[a].size(); i++) {
      b = w[a][i];
      if (del[b]) legs[a] = legs[a] + 1;
    }
  }
  for (int a = 1; a <= n; ++a)
    if (!del[a]) {
      int cnt = 0;
      for (int i = 0; i < w[a].size(); i++) {
        b = w[a][i];
        if (!del[b] && (legs[b] > 2 || w[b].size() != legs[b] + 1)) ++cnt;
      }
      if (cnt > 2) {
        puts("No");
        return 0;
      }
    }
  puts("Yes");
  return 0;
}
