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;
bool vis[100005];
vector<long long> v[100005];
bool flag;
long long cut;
long long dfs(long long ind) {
  long long i;
  if (!vis[ind]) {
    vis[ind] = true;
    long long sz = v[ind].size();
    if (ind == cut) sz--;
    if (sz >= 3) return false;
    for (i = 0; i < v[ind].size(); i++) {
      if (dfs(v[ind][i]) == false) return false;
    }
  }
  return true;
}
int main() {
  long long i, n, x, y, c, j;
  scanf("%lld", &n);
  for (i = 1; i <= n - 1; i++) {
    scanf("%lld %lld", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (i = 1; i <= n; i++) {
    if (v[i].size() <= 2) continue;
    c = 0;
    memset(vis, false, sizeof(vis));
    vis[i] = true;
    for (j = 0; j < v[i].size(); j++) {
      cut = v[i][j];
      if (!dfs(v[i][j])) c++;
    }
    if (c >= 3) {
      printf("NO");
      return 0;
    }
  }
  printf("YES");
  return 0;
}
