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 int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int head[100005], nxt[100005 * 2], to[100005 * 2], cnt;
int n, sz;
int degree[100005];
bool is_lef[100005];
int legs[100005];
inline void add_edge(int a, int b) {
  to[++cnt] = b;
  nxt[cnt] = head[a];
  head[a] = cnt;
  to[++cnt] = a;
  nxt[cnt] = head[b];
  head[b] = cnt;
}
void dfs(int x, int f) {
  is_lef[x] = 1;
  for (int i = head[x]; i; i = nxt[i])
    if (to[i] != f)
      if (degree[to[i]] <= 2) dfs(to[i], x);
}
int main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int a = read(), b = read();
    add_edge(a, b);
    degree[a]++;
    degree[b]++;
  }
  for (int i = 1; i <= n; i++)
    if (degree[i] == 1) dfs(i, 0);
  for (int i = 1; i <= n; i++)
    for (int j = head[i]; j; j = nxt[j])
      if (is_lef[to[j]]) legs[i]++;
  for (int i = 1; i <= n; i++) {
    if (is_lef[i]) continue;
    sz = 0;
    for (int j = head[i]; j; j = nxt[j])
      if (!is_lef[to[j]] && degree[to[j]] - min(legs[to[j]], 2) >= 2) sz++;
    if (sz > 2) {
      puts("No");
      return 0;
    }
  }
  puts("Yes");
  return 0;
}
