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 n, a[1000000][2], last[200000], x, y, f[200000][10], kk;
void doit(int x, int y) {
  a[++kk][0] = last[x];
  a[kk][1] = y;
  last[x] = kk;
}
int init(int x) {
  if (x == 0) return 0;
  if (f[x][5] > 0 || f[x][4] > 0 || f[x][3] > 2) return 5;
  if (f[x][3] > 1) return 4;
  if (f[x][3] > 0 || f[x][2] > 0 || f[x][1] > 2) return 3;
  if (f[x][1] > 1) return 2;
  return 1;
}
void dfs1(int d, int fa) {
  for (int i = last[d]; i; i = a[i][0])
    if (a[i][1] != fa) dfs1(a[i][1], d), f[d][init(a[i][1])]++;
}
void dfs2(int d, int fa) {
  int k = init(d);
  f[fa][k]--;
  f[d][init(fa)]++;
  if (init(d) < 5) {
    cout << "Yes" << endl;
    exit(0);
  }
  f[fa][k]++;
  for (int i = last[d]; i; i = a[i][0])
    if (a[i][1] != fa) dfs2(a[i][1], d);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; i++) cin >> x >> y, doit(x, y), doit(y, x);
  dfs1(1, 0);
  dfs2(1, 0);
  cout << "No" << endl;
}
