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;
template <typename T>
T Abs(T first) {
  return (first < 0 ? -first : first);
}
template <typename T>
T Sqr(T first) {
  return (first * first);
}
string plural(string s) {
  return (int((s).size()) && s[int((s).size()) - 1] == 'x' ? s + "en"
                                                           : s + "s");
}
const int INF = (int)1e9;
const long double EPS = 1e-9;
const long double PI = acos(-1.0);
bool Read(int &first) {
  char c, r = 0, n = 0;
  first = 0;
  for (;;) {
    c = getchar();
    if ((c < 0) && (!r)) return (0);
    if ((c == '-') && (!r))
      n = 1;
    else if ((c >= '0') && (c <= '9'))
      first = first * 10 + c - '0', r = 1;
    else if (r)
      break;
  }
  if (n) first = -first;
  return (1);
}
int main() {
  if (0) freopen("in.txt", "r", stdin);
  int N;
  int i, j, k, p, pp;
  static int deg[100000], deg2[100000];
  static bool vis[100000];
  static vector<int> con[100000];
  Read(N);
  if (N == 1) goto Yes;
  for (i = 0; i < N - 1; i++) {
    Read(j), Read(k);
    j--, k--;
    deg[j]++;
    deg[k]++;
    con[j].push_back(k);
    con[k].push_back(j);
  }
  for (i = 0; i < N; i++)
    if (deg[i] == 1) {
      vis[i] = 1;
      p = i;
      j = con[i][0];
      while (deg[j] < 3) {
        pp = j;
        vis[j] = 1;
        if (deg[j] == 1) goto Yes;
        if (con[j][0] == p)
          j = con[j][1];
        else
          j = con[j][0];
        p = pp;
      }
      deg2[j]++;
    }
  for (i = 0; i < N; i++)
    if ((deg2[i] == 2) && (deg[i] == 3)) vis[i] = 1;
  for (i = 0; i < N; i++)
    if (!vis[i]) {
      k = 0;
      for (j = 0; j < int((con[i]).size()); j++)
        if (!vis[con[i][j]]) k++;
      if (k > 2) goto No;
    }
Yes:;
  printf("Yes\n");
  return (0);
No:;
  printf("No\n");
  return (0);
}
