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;
template <class T>
inline bool getInt(T &x) {
  char c = 0;
  T sign = 1;
  if ((c = getchar()) == EOF) {
    return false;
  }
  while ((c < '0' || c > '9') && c != '-') {
    c = getchar();
  };
  if (c == '-') {
    sign = -1;
    c = getchar();
  }
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') {
    x = x * 10 + c - '0';
  }
  x *= sign;
  return true;
}
int n;
int rnk[100005];
bool del[100005];
vector<int> e[100005];
void dfs(int u, int pre) {
  if (e[u].size() <= 2) {
    del[u] = true;
    for (size_t i = 0; i < e[u].size(); i++) {
      int v = e[u][i];
      if (v != pre) {
        dfs(v, u);
      }
    }
  }
}
int main() {
  int u, v;
  getInt(n);
  for (int i = 1; i < n; i++) {
    getInt(u);
    getInt(v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  for (int u = 1; u <= n; u++) {
    if (e[u].size() == 1) {
      dfs(u, 0);
    }
  }
  for (int u = 1; u <= n; u++) {
    for (size_t i = 0; i < e[u].size(); i++) {
      int v = e[u][i];
      if (del[v]) {
        rnk[u] = min(rnk[u] + 1, 2);
      }
    }
  }
  for (int u = 1; u <= n; u++) {
    if (!del[u]) {
      int cnt = 0;
      for (size_t i = 0; i < e[u].size(); i++) {
        int v = e[u][i];
        if (!del[v] && e[v].size() - rnk[v] > 1) {
          cnt++;
        }
      }
      if (cnt > 2) {
        puts("No");
        return 0;
      }
    }
  }
  puts("Yes");
}
