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;
struct Ttree {
  static const int maxn = 100010;
  struct Tedge {
    int son;
    Tedge *pre;
    void add(int s_, Tedge *p_) { son = s_, pre = p_; }
  } edge[maxn << 1], *pos[maxn];
  int tot;
  void add(int a, int b) { edge[++tot].add(b, pos[a]), pos[a] = edge + tot; }
  void clear() {
    tot = 0;
    memset(pos, 0, sizeof(pos));
  }
} t;
const int maxn = 100010;
int deg[maxn];
int n, root;
void init() {
  scanf("%d", &n), t.clear();
  memset(deg, 0, sizeof(deg));
  for (int i = 1, a, b; i <= n - 1; ++i) {
    scanf("%d%d", &a, &b);
    t.add(a, b), t.add(b, a);
    ++deg[a], ++deg[b];
  }
  root = -1;
  for (int i = 1; i <= n; ++i)
    if (deg[i] >= 3) {
      root = i;
      break;
    }
}
int type[maxn];
bool dp(int x, int fa) {
  for (Ttree::Tedge *i = t.pos[x]; i; i = i->pre)
    if (i->son != fa && !dp(i->son, x)) return false;
  if (x == root) {
    int sum = 0;
    for (Ttree::Tedge *i = t.pos[x]; i; i = i->pre)
      if (i->son != fa && type[i->son] == -1) ++sum;
    return sum <= 2;
  }
  switch (deg[x]) {
    case 1:
      type[x] = 1;
      return true;
    case 2:
      for (Ttree::Tedge *i = t.pos[x]; i; i = i->pre)
        if (i->son != fa) {
          type[x] = (type[i->son] == 1) ? 1 : -1;
          break;
        }
      return true;
    case 3:
      type[x] = 2;
      for (Ttree::Tedge *i = t.pos[x]; i; i = i->pre)
        if (i->son != fa && type[i->son] != 1) {
          type[x] = -1;
          break;
        }
      for (Ttree::Tedge *i = t.pos[x]; i; i = i->pre)
        if (i->son != fa && type[i->son] != -1) return true;
      return false;
    default:
      type[x] = -1;
      int sum = 0;
      for (Ttree::Tedge *i = t.pos[x]; i; i = i->pre)
        if (i->son != fa && type[i->son] == -1) ++sum;
      return sum <= 1;
  }
}
int main() {
  init();
  if (root == -1) {
    puts("Yes");
    return 0;
  }
  if (deg[root] == 3) {
    dp(root, 0);
    for (Ttree::Tedge *i = t.pos[root]; i; i = i->pre)
      if (type[i->son] == -1) root = i->son;
  }
  if (dp(root, 0))
    puts("Yes");
  else
    puts("No");
  return 0;
}
