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;
long long n, sz[100005], lg[100005];
bool del[100005];
vector<long long> e[100005];
void dfs(long long nw, long long sn) {
  if (sz[nw] <= 2) {
    del[nw] = 1;
    for (int i = 0; i < sz[nw]; i++)
      if (e[nw][i] != sn) dfs(e[nw][i], nw);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) sz[i] = e[i].size();
  for (int i = 1; i <= n; i++) {
    if (sz[i] == 1) dfs(i, 0);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < sz[i]; j++) {
      if (del[e[i][j]]) lg[i]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!del[i]) {
      long long cnt = 0;
      for (int j = 0; j < sz[i]; j++) {
        if (sz[e[i][j]] - min(lg[e[i][j]], 2ll) > 1 && (!del[i])) ++cnt;
        if (cnt > 2) return printf("No"), 0;
      }
    }
  }
  return printf("Yes"), 0;
}
