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 = 110000;
vector<int> nxt[Maxn];
bool del[Maxn];
int leg[Maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    nxt[x].push_back(y);
    nxt[y].push_back(x);
  }
  queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (nxt[i].size() == 1) q.push(i);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    del[cur] = true;
    for (int j = 0; j < nxt[cur].size(); ++j) {
      int y = nxt[cur][j];
      if (!del[y] && nxt[y].size() < 3) {
        q.push(y);
        del[y] = true;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < nxt[i].size(); ++j) {
      int y = nxt[i][j];
      if (del[y]) leg[i] = min(leg[i] + 1, 2);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (del[i]) continue;
    int s = 0;
    for (int j = 0; j < nxt[i].size(); ++j) {
      int y = nxt[i][j];
      if (!del[y] && (int)nxt[y].size() - leg[y] > 1) s++;
    }
    if (s > 2) {
      printf("No\n");
      return 0;
    }
  }
  printf("Yes\n");
  return 0;
}
