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 n, deg[100100], a, b, del[100100], legs[100100];
vector<vector<int> > g;
void dfs(int u) {
  del[u] = 1;
  for (int j = 0; j < (int)g[u].size(); j++) {
    int v = g[u][j];
    if (deg[v] < 3 && !del[v]) {
      dfs(v);
    } else if (deg[v] >= 3) {
      legs[v]++;
    }
  }
}
int countTN(int u) {
  int cnt = 0;
  for (int j = 0; j < (int)g[u].size(); j++) {
    int v = g[u][j];
    if (!del[v] && deg[v] - min(2, legs[v]) > 1) {
      cnt++;
    }
  }
  return cnt;
}
int main() {
  scanf("%d", &n);
  g.assign(n + 1, vector<int>());
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &a, &b);
    deg[a]++, deg[b]++;
    g[a].push_back(b), g[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1 && !del[i]) {
      dfs(i);
    }
  }
  bool ans = true;
  for (int i = 1; i <= n; i++) {
    if (!del[i] && countTN(i) > 2) {
      ans = false;
      break;
    }
  }
  printf("%s\n", (ans) ? "Yes" : "No");
}
