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 MOD(1000000007);
const int INF((1 << 30) - 1);
const int MAXN(100005);
vector<int> a[MAXN];
int visited[MAXN], num[MAXN], cnt;
void dfs(int node, int c, int d, int l) {
  visited[node] = c;
  if (a[node].size() > 3 || (a[node].size() == 3 && l != 1)) {
    if (num[d] != c) {
      cnt++;
      num[d] = c;
    }
    return;
  }
  for (int i = 0; i < a[node].size(); i++) {
    int u = a[node][i];
    if (visited[u] != c) dfs(u, c, d, l + 1);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (a[i].size() >= 3) {
      cnt = 0;
      visited[i] = i;
      for (int j = 0; j < a[i].size(); j++) dfs(a[i][j], i, j + 1, 1);
      if (cnt > 2) {
        printf("No");
        return 0;
      }
    }
  }
  printf("Yes");
}
