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;
const int N = 1e5 + 10;
inline int rd() {
  int x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  if ((x *= w) == -1) exit(0);
  return x;
}
int to[N << 1], nt[N << 1], hd[N], dg[N], tot = 1;
inline void add(int x, int y) {
  ++tot, to[tot] = y, nt[tot] = hd[x], hd[x] = tot, ++dg[x];
  ++tot, to[tot] = x, nt[tot] = hd[y], hd[y] = tot, ++dg[y];
}
int n, sht[N];
bool ban[N];
int main() {
  n = rd();
  for (int i = 1; i < n; ++i) add(rd(), rd());
  queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (dg[i] == 1) ban[i] = 1, q.push(i);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = hd[x]; i; i = nt[i])
      if (!ban[to[i]] && dg[to[i]] <= 2) ban[to[i]] = 1, q.push(to[i]);
  }
  for (int i = 1; i <= n; ++i)
    if (!ban[i]) {
      for (int j = hd[i]; j; j = nt[j]) sht[i] += ban[to[j]];
      sht[i] = (dg[i] - min(sht[i], 2) > 1);
    }
  bool ok = 1;
  for (int i = 1; i <= n && ok; ++i)
    if (!ban[i]) {
      int cn = 0;
      for (int j = hd[i]; j; j = nt[j]) cn += sht[to[j]];
      ok = cn <= 2;
    }
  ok ? puts("Yes") : puts("No");
  return 0;
}
