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;
vector<int> nodos[100010];
int patas[100010];
bool eliminado[100010];
int main() {
  memset(patas, 0, sizeof(patas));
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    nodos[a].push_back(b);
    nodos[b].push_back(a);
  }
  for (int i = 1; i < n + 1; i++)
    if (nodos[i].size() == 1) {
      int padre = nodos[i][0], hijo = i;
      eliminado[hijo] = true;
      while (nodos[padre].size() == 2) {
        int temp = hijo;
        if (nodos[padre][0] != hijo)
          temp = nodos[padre][0];
        else
          temp = nodos[padre][1];
        hijo = padre;
        padre = temp;
        eliminado[hijo] = true;
      }
      padre[patas]++;
    }
  for (int i = 1; i < n + 1; i++)
    if (!eliminado[i] && nodos[i].size() == 3 && patas[i] == 2) {
      eliminado[i] = true;
      for (int j = 0; j < 3; j++)
        if (!eliminado[nodos[i][j]]) patas[nodos[i][j]]++;
    }
  for (int i = 1; i < n + 1; i++)
    if (!eliminado[i] && nodos[i].size() - patas[i] > 2) {
      printf("No\n");
      return 0;
    }
  printf("Yes\n");
}
