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 N = 100005;
int n, numberOfPaths[N];
vector<int> graph[N];
bool deleted[N];
void propagate(int u, int p = -1) {
  if ((int)graph[u].size() <= 2) {
    deleted[u] = true;
    for (int &v : graph[u]) {
      if (v == p) continue;
      propagate(v, u);
    }
  }
}
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--, v--;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  for (int i = 0; i < n; i++) {
    if ((int)graph[i].size() == 1) {
      propagate(i);
    }
  }
  for (int i = 0; i < n; i++) {
    if (deleted[i]) continue;
    for (int v : graph[i]) {
      if (!deleted[v]) continue;
      numberOfPaths[i]++;
      if (numberOfPaths[i] > 2) numberOfPaths[i] = 2;
    }
  }
  bool ans = true;
  for (int i = 0; i < n; i++) {
    if (deleted[i]) continue;
    int mainPathNeighbors = 0;
    for (int v : graph[i]) {
      if (deleted[v]) continue;
      if ((int)graph[v].size() - numberOfPaths[v] > 1) mainPathNeighbors++;
    }
    if (mainPathNeighbors > 2) ans = false;
  }
  if (ans)
    puts("Yes");
  else
    puts("No");
  return 0;
}
