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 MN = 1e5 + 100;
vector<int> v[MN];
int ok = 1;
bool used[MN];
int f(int x, int root) {
  used[x] = 1;
  int ret = 1;
  int cnt3 = 0;
  int cnt1 = 0;
  for (int i = 0; i < v[x].size(); ++i) {
    int y = v[x][i];
    if (!used[y]) {
      int z = f(y, root);
      if (z == 1) ++cnt1;
      if (z == 2) {
        ret = max(ret, 3);
      }
      if (z == 3) {
        ret = max(ret, 3);
        ++cnt3;
      }
    }
  }
  if (x == root) {
    if (cnt3 > 2) ok = 0;
    return 0;
  }
  if (ret == 1) {
    if (cnt1 <= 1) return 1;
    if (cnt1 == 2) return 2;
    return 3;
  }
  if (cnt3 > 1) ok = 0;
  return 3;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  int root = -1;
  for (int i = 1; i <= n; ++i) {
    if (v[i].size() >= 3) {
      root = i;
    }
  }
  if (root == -1) {
    cout << "Yes\n";
    return 0;
  }
  f(root, root);
  int lol = ok;
  ok = 1;
  for (int i = 0; i < v[root].size(); ++i) {
    if (v[v[root][i]].size() >= 3) {
      memset(used, 0, sizeof used);
      f(v[root][i], v[root][i]);
      lol = max(lol, ok);
      break;
    }
  }
  if (lol) {
    cout << "Yes\n";
  } else {
    cout << "No\n";
  }
}
