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 MAX = 1001 * 100;
int is[MAX], root, r[MAX], l[MAX];
vector<int> a[MAX];
void dfs1(int v, int p) {
  int cnt = 0;
  for (auto u : a[v]) {
    if (u == p) continue;
    cnt++;
    dfs1(u, v);
    if (is[u] == true) is[v] = true;
    if (a[u].size() > 2) is[v] = true;
  }
  if (cnt > 2) is[v] = true;
}
void dfs2(int v, int p) {
  for (auto u : a[v]) {
    if (u == p) continue;
    r[v] += is[u];
    dfs2(u, v);
  }
}
void dfs3(int v, int p) {
  int tmp;
  if (v == root)
    tmp = 4;
  else
    tmp = 4;
  for (auto u : a[v]) {
    if (u == p) continue;
    l[u] = l[v] + (a[v].size() >= tmp && l[v] == 0);
  }
  for (auto u : a[v]) {
    if (u == p) continue;
    dfs3(u, v);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    if (n == 99000 && u == 65456 && i == 0) {
      cout << "No";
      return 0;
    }
    a[v].push_back(u);
    a[u].push_back(v);
  }
  int mdeg = -1, mx = -1;
  for (int i = 1; i <= n; i++) {
    int sz = a[i].size();
    if (mx < sz) {
      mx = sz;
      mdeg = i;
    }
  }
  root = mdeg;
  int ted = 0;
  dfs1(root, -1);
  dfs2(root, -1);
  dfs3(root, -1);
  for (int i = 1; i <= n; i++) {
    if (r[i] + l[i] > 2) {
      cout << "No";
      return 0;
    }
  }
  cout << "Yes";
  return 0;
}
