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 = 1e5 + 5;
bool used[MAXN];
vector<int> g[MAXN];
int dp[MAXN];
void dfs(int v) {
  if ((int)g[v].size() <= 2) {
    used[v] = 1;
    for (int j = 0; j < (int)g[v].size(); j++) {
      int to = g[v][j];
      if (!used[to]) {
        dfs(to);
      }
    }
  }
}
void solve() {
  int n;
  cin >> n;
  int v, to;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &v, &to);
    g[v].push_back(to);
    g[to].push_back(v);
  }
  for (int i = 1; i <= n; i++) {
    if ((int)g[i].size() == 1 && !used[i]) {
      dfs(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < (int)g[i].size(); j++) {
      int to = g[i][j];
      if (used[to]) {
        dp[i] = min(2, dp[i] + 1);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (used[i]) continue;
    int cur = 0;
    for (int j = 0; j < (int)g[i].size(); j++) {
      int to = g[i][j];
      if (!used[i] && (int)g[to].size() - dp[to] > 1) {
        cur++;
      }
    }
    if (cur > 2) {
      printf("No");
      return;
    }
  }
  printf("Yes");
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
