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;
const int N = 112345;
vector<int> v[N];
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
int cnt[N];
bool used[N];
bool t[N];
bool never[N];
int main() {
  int n, m, i, j, x, y;
  n = read();
  for (i = 2; i <= n; i++) {
    x = read();
    y = read();
    v[x].push_back(y);
    cnt[x]++;
    cnt[y]++;
    v[y].push_back(x);
  }
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (cnt[i] == 1) {
      q.push(i);
      used[i] = 1;
    }
  }
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = 0; i < v[x].size(); i++) {
      int to = v[x][i];
      if (!used[to]) {
        if (cnt[to] <= 2 && !t[x] && !never[to]) {
          used[to] = 1;
          q.push(to);
        } else if (cnt[to] == 3 && !t[x] && !never[to]) {
          t[to] = 1;
        } else {
          never[to] = 1;
        }
        cnt[to]--;
      }
    }
  }
  for (i = 1; i <= n; i++) {
    if (cnt[i] >= 3) {
      puts("no");
      return 0;
    }
  }
  puts("yes");
}
