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;
inline int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, num[100005], flag[100005];
vector<int> a[100005];
void dfs(int u, int last) {
  if (a[u].size() <= 2) {
    flag[u] = 1;
    for (int i = 0; i <= a[u].size() - 1; ++i)
      if (a[u][i] != last) dfs(a[u][i], u);
  }
}
int main() {
  n = read();
  for (int i = 2; i <= n; ++i) {
    int u = read(), v = read();
    a[u].push_back(v);
    a[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i)
    if (a[i].size() == 1) dfs(i, 0);
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < a[i].size(); ++j) num[i] += flag[a[i][j]];
  for (int i = 1; i <= n; ++i)
    if (!flag[i]) {
      int cnt = 0;
      for (int j = 0; j < a[i].size(); ++j) {
        int v = a[i][j];
        if (!flag[v] && a[v].size() > min(num[v], 2) + 1) ++cnt;
      }
      if (cnt > 2) return puts("No") & 0;
    }
  puts("Yes");
}
