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;
char _;
const int maxN = 100 * 1000 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
const int base = 701;
const int maxL = 20;
vector<int> a[maxN];
int sz[maxN];
bool fl;
void dfs(int v, int par = -1) {
  if (a[v].size() == 3)
    sz[v] = 1;
  else if (a[v].size() > 3)
    sz[v] = 2;
  int ted = 0;
  for (auto u : a[v]) {
    if (u == par) continue;
    dfs(u, v);
    if (sz[u] - (a[u].size() > 2)) ted++;
    sz[v] += sz[u];
  }
  if (par == -1 && ted > 2) fl = 1;
  if (par > -1 && ted > 1) fl = 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    v--;
    u--;
    a[v].push_back(u);
    a[u].push_back(v);
  }
  if (n <= 4) {
    cout << "Yes" << endl;
    return 0;
  }
  int root = 0;
  int mx = a[0].size();
  for (int i = 1; i < n; i++)
    if (a[i].size() > mx) mx = a[i].size(), root = i;
  if (mx <= 2) {
    cout << "Yes" << endl;
    return 0;
  }
  dfs(root);
  if (fl == 0) {
    cout << "Yes" << endl;
    return 0;
  }
  if (mx == 3) {
    vector<int> ver;
    for (int i = 0; i < n; i++)
      if (a[i].size() == 3) ver.push_back(i);
    random_shuffle(ver.begin(), ver.end());
    for (int i = 0; i < min((int)ver.size(), 300); i++) {
      memset(sz, 0, sizeof sz);
      fl = 0;
      dfs(ver[i]);
      if (fl == 0) {
        cout << "Yes" << endl;
        return 0;
      }
    }
  }
  cout << "No" << endl;
  return 0;
}
