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;
int legs[100005];
vector<int> E[100005];
vector<int> leafs;
bool vis[100005];
bool deleted[100005];
int deg(int u) { return E[u].size(); }
void findL(int u) {
  vis[u] = true;
  if (E[u].size() == 1) leafs.push_back(u);
  for (int i = 0, v; i < E[u].size(); i++) {
    v = E[u][i];
    if (vis[v]) continue;
    findL(v);
  }
}
void del(int u) {
  if (deg(u) > 2) {
    legs[u]++;
    return;
  }
  deleted[u] = true;
  for (int i = 0, v; i < deg(u); i++) {
    v = E[u][i];
    if (deleted[v]) continue;
    del(v);
  }
}
int main() {
  scanf("%d", &n);
  int a, b;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &b);
    a--;
    b--;
    E[a].push_back(b);
    E[b].push_back(a);
  }
  findL(0);
  for (int i = 0; i < leafs.size(); i++) {
    del(leafs[i]);
  }
  bool pos = true;
  for (int u = 0; u < n && pos; u++) {
    int cnt = 0;
    if (deleted[u]) continue;
    for (int i = 0, v; i < deg(u); i++) {
      v = E[u][i];
      if (deleted[v]) continue;
      if (deg(v) - legs[v] == 1) {
        if (legs[v] != 2) cnt++;
      } else {
        cnt++;
      }
    }
    if (cnt > 2) pos = false;
  }
  if (pos)
    puts("Yes");
  else
    puts("No");
}
