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>
#pragma warning(disable : 4996)
using namespace std;
int N;
vector<int> E[100000];
int side[100000];
int chain[100000];
int line[100000];
int V[100000];
bool ans = true;
int rand32() { return (rand() << 16) + rand(); }
void dfs(int n) {
  V[n] = 1;
  vector<int> ch;
  for (int t : E[n])
    if (!V[t]) {
      dfs(t);
      ch.push_back(t);
    }
  if (ch.empty()) {
    line[n] = chain[n] = side[n] = 1;
    return;
  }
  if (ch.size() == 1 && line[ch[0]]) line[n] = 1;
  if (ch.size() == 2 && line[ch[0]] && line[ch[1]]) chain[n] = 1;
  chain[n] |= line[n];
  int cnt = 0;
  for (int p : ch) {
    if (chain[p]) cnt++;
  }
  if (cnt + 1 >= ch.size()) {
    side[n] = 1;
    for (int p : ch) {
      if (!chain[p] && !side[p]) side[n] = 0;
    }
  }
  bool tmp = true;
  if (cnt + 2 >= ch.size()) {
    int sp;
    for (int p : ch) {
      if (!chain[p] && !side[p]) {
        tmp = false;
        sp = p;
      }
    }
    if (tmp == true) return;
    if (ch.size() > 3) {
      ans = false;
      return;
    }
    for (int p : ch) {
      if (p != sp && !line[p]) ans = false;
    }
  } else
    ans = false;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b), --a, --b;
    E[a].push_back(b), E[b].push_back(a);
  }
  for (int cc = 20; cc >= 0; cc--) {
    for (int i = 0; i < N; i++) V[i] = 0, line[i] = side[i] = chain[i] = 0;
    int n = rand32() % N;
    dfs(n);
    if (ans == true) {
      puts("Yes");
      return 0;
    }
  }
  puts("No");
  return 0;
}
