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;
int deg[100009], cnt[100009];
vector<int> e[100009];
void dfs(int u, int p, int t) {
  if (t > 0 && deg[u] > 2) {
    cnt[u]++;
    cnt[t]++;
  }
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (v == p) continue;
    if (deg[u] > 2)
      dfs(v, u, u);
    else
      dfs(v, u, t);
  }
}
int main() {
  int n, i, j, u, v;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  dfs(1, 0, 0);
  for (i = 1; i <= n; i++) {
    if (deg[i] > 2) {
      u = 0;
      for (int j = 0; j < e[i].size(); j++) {
        v = e[i][j];
        if (deg[v] == 3 && cnt[v] == 1) u++;
      }
      if (cnt[i] > u + 2) {
        printf("No\n");
        return 0;
      }
    }
  }
  printf("Yes\n");
  return 0;
}
