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 Readint() {
  int ret = 0;
  int f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f *= -1;
  } while (ch >= 0 && (ch < '0' || ch > '9'));
  while ('0' <= ch && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * f;
}
int trans[10][10] = {
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 2, 2},
    {0, 0, 0, 0, 0, 0, 0, 4, 4, 4}, {0, 0, 0, 0, 0, 4, 4, 5, 5, 5},
    {0, 0, 0, 0, 0, 5, 5, 6, 6, 6}, {0, 0, 0, 0, 1, 5, 5, 6, 6, 6},
    {0, 0, 1, 1, 2, 5, 5, 6, 7, 7}, {0, 0, 2, 2, 3, 5, 5, 6, 8, 8}};
void open() {
  freopen("C.in", "r", stdin);
  freopen("C.out", "w", stdout);
}
void close() {
  fclose(stdin);
  fclose(stdout);
}
const int MAXN = 1E5 + 10;
int n;
int to[MAXN * 2];
int nxt[MAXN * 2];
int head[MAXN];
int tot;
inline void add(const int &x, const int &y) {
  tot++;
  to[tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
}
void init() {
  n = Readint();
  int x, y;
  for (int i = 2, _END_ = n; i <= _END_; i++) {
    x = Readint(), y = Readint();
    add(x, y);
    add(y, x);
  }
}
int dfs(int p, int fa) {
  int ret = 9;
  for (int i = head[p]; i; i = nxt[i])
    if (to[i] != fa) ret = trans[ret][dfs(to[i], p)];
  return ret;
}
int main() {
  int _ = 0;
  init();
  if (dfs(1, 0))
    printf("Yes\n");
  else
    printf("No\n");
  close();
  return 0;
}
