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, dp[100005];
int ts[10][10] = {
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 2, 2},
    {0, 0, 0, 0, 0, 0, 0, 4, 4, 4}, {0, 0, 0, 0, 0, 4, 4, 5, 5, 5},
    {0, 0, 0, 0, 0, 5, 5, 6, 6, 6}, {0, 0, 0, 0, 1, 5, 5, 6, 6, 6},
    {0, 0, 1, 1, 2, 5, 5, 6, 7, 7}, {0, 0, 2, 2, 3, 5, 5, 6, 8, 8}};
vector<int> V[100005];
void dfs(int id, int prev) {
  dp[id] = 9;
  for (int i = 0; i < V[id].size(); i++) {
    int to = V[id][i];
    if (to == prev) continue;
    dfs(to, id);
    dp[id] = ts[dp[id]][dp[to]];
  }
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    V[a].push_back(b);
    V[b].push_back(a);
  }
  dfs(1, -1);
  if (dp[1] == 0)
    printf("No\n");
  else
    printf("Yes\n");
  return 0;
}
