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 maxn = 100101;
int pre[maxn], ind[maxn], cnt[maxn], f[maxn];
vector<int> g[maxn], q;
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    g[l].push_back(r);
    g[r].push_back(l);
    ind[l]++;
    ind[r]++;
  }
  q.push_back(1);
  for (int i = 0; i < q.size(); ++i) {
    int now = q[i];
    for (int j = 0; j < g[now].size(); ++j)
      if (g[now][j] != pre[now]) {
        pre[g[now][j]] = now;
        q.push_back(g[now][j]);
      }
  }
  bool flag = 1;
  for (int i = (int)q.size() - 1; i >= 0; --i) {
    int now = q[i];
    int k = cnt[now] + (ind[now] > 3);
    if (k) f[pre[now]]++;
    if (ind[now] > 2) cnt[now]++;
    cnt[pre[now]] += cnt[now];
  }
  int sum = cnt[1];
  for (int i = 0; i < q.size(); ++i) {
    int now = q[i];
    for (int j = 0; j < g[now].size(); ++j)
      if (g[now][j] != pre[now]) {
        int k = sum - cnt[g[now][j]] - (ind[now] > 2) + (ind[now] > 3);
        if (k) f[g[now][j]]++;
      }
  }
  for (int i = 1; i <= n; ++i)
    if (f[i] > 2) flag = 0;
  if (flag)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
