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<int> g[(1 << 17)];
int n, a, b, d[(1 << 17)], l[(1 << 17)], C;
void dfs(int nd, int p) {
  if (g[nd].size() < 3) {
    d[nd] = 1;
    for (int i : g[nd])
      if (i != p) dfs(i, nd);
  }
}
int main(void) {
  scanf("%d", &n);
  for (int i(1); i < n; ++i)
    scanf("%d%d", &a, &b), g[--a].push_back(--b), g[b].push_back(a);
  for (int i(0); i < n; ++i)
    if (g[i].size() == 1) dfs(i, -1);
  for (int i(0); i < n; ++i)
    for (int j : g[i])
      if (d[j]) l[i] = min(2, l[i] + 1);
  for (int i(0), c(0); i < n; ++i, c = 0)
    if (!d[i]) {
      for (int j : g[i])
        if (!d[j] && g[j].size() - l[j] > 1) ++c;
      C |= c > 2;
    }
  printf("%s\n", C ? "No" : "Yes");
  return 0;
}
