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;
int tot, head[1200000], to[1200000], nx[1200000], n, gen, du[1200000];
bool vis[1200000];
void add(int a, int b) {
  tot++;
  nx[tot] = head[a];
  to[tot] = b;
  head[a] = tot;
  return;
}
int dfs(int x) {
  int sxd[10] = {0};
  bool flag = false;
  for (int i = head[x]; i; i = nx[i]) {
    int at = to[i];
    if (vis[at] == false) {
      vis[at] = true;
      sxd[dfs(at)]++;
      flag = 1;
    }
  }
  if (flag == false)
    return 1;
  else if (sxd[1] == 1 && sxd[2] == 0 && sxd[3] == 0 && sxd[4] == 0 &&
           sxd[5] == 0)
    return 1;
  else if (sxd[1] == 2 && sxd[2] == 0 && sxd[3] == 0 && sxd[4] == 0 &&
           sxd[5] == 0)
    return 2;
  else if (((sxd[1] >= 3 && sxd[3] <= 1) || (sxd[2] >= 1 && sxd[3] <= 1) ||
            sxd[3] == 1) &&
           sxd[4] == 0 && sxd[5] == 0)
    return 3;
  else if ((sxd[3] == 2 && sxd[4] == 0 && sxd[5] == 0))
    return 4;
  else
    return 5;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    add(x, y);
    add(y, x);
    du[x]++;
    du[y]++;
  }
  for (int i = 1; i <= 50; i++) {
    memset(vis, 0, sizeof vis);
    gen = rand() % n + 1;
    vis[gen] = 1;
    if (dfs(gen) < 5) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
  return 0;
}
