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 mx = 1e5 + 10;
int main() {
  int n;
  scanf("%d", &n);
  vector<vector<int> > gr(n);
  int deg[mx] = {};
  bool gone[mx] = {};
  for (int i = (0); i < int(n - 1); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    gr[a].push_back(b);
    gr[b].push_back(a);
    ++deg[a];
    ++deg[b];
  }
  stack<int> junk;
  for (int i = (0); i < int(n); ++i)
    if (deg[i] == 1 && deg[gr[i][0]] <= 2) junk.push(i);
  while (!junk.empty()) {
    int j = junk.top();
    junk.pop();
    gone[j] = 1;
    for (int v : gr[j])
      if (!gone[v]) {
        if (--deg[v] == 1) {
          for (int vv : gr[v])
            if (!gone[vv]) {
              if (deg[vv] <= 2) junk.push(v);
              break;
            }
        }
        break;
      }
  }
  int leafDeg[mx] = {};
  for (int i = (0); i < int(n); ++i)
    if (!gone[i] && deg[i] == 1)
      for (int v : gr[i]) ++leafDeg[v];
  for (int i = (0); i < int(n); ++i) {
    if (gone[i]) continue;
    int cnt = 0;
    for (int v : gr[i])
      if (!gone[v] && deg[v] - min(2, leafDeg[v]) > 1) ++cnt;
    if (cnt > 2) {
      puts("No");
      return 0;
    }
  }
  puts("Yes");
}
