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;
inline void read(int &x) {
  char c = getchar();
  x = 0;
  while (c > '9' || c < '0') c = getchar();
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
}
const int N = 100005;
int b[N], d[N], red[N], x, y, to[N * 2], pr[N * 2], la[N], cnt, n;
inline void add(int x, int y) {
  to[++cnt] = y;
  pr[cnt] = la[x];
  la[x] = cnt;
}
void dfs(int x, int fa) {
  if (d[x] <= 2) {
    b[x] = 1;
    for (int i = la[x]; i; i = pr[i])
      if (to[i] != fa) dfs(to[i], x);
  }
}
int main() {
  read(n);
  for (int i = 1; i < n; ++i) {
    read(x);
    read(y);
    add(x, y);
    add(y, x);
    ++d[x];
    ++d[y];
  }
  for (int i = 1; i <= n; ++i)
    if (d[i] == 1) dfs(i, 0);
  for (int i = 1; i <= n; ++i)
    if (d[i] >= 3) {
      int sum = 0;
      for (int j = la[i]; j; j = pr[j])
        if (b[to[j]]) ++sum;
      red[i] = d[i] - sum;
    }
  for (int i = 1; i <= n; ++i)
    if (d[i] == 3 && red[i] == 1) b[i] = 1;
  for (int i = 1; i <= n; ++i)
    if (!b[i]) {
      int sum = 0;
      for (int j = la[i]; j; j = pr[j])
        if (!b[to[j]]) ++sum;
      if (sum > 2) {
        puts("No");
        return 0;
      }
    }
  puts("YES");
  return 0;
}
