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 inf = 0x3f3f3f3f;
const int maxn = 100010;
vector<int> g[maxn];
int n;
int root = -1;
bool flag = true;
int dfs(int x, int f) {
  if (g[x].size() == 1) return 1;
  int a = 0, b = 0, c = 0;
  for (int i = 0; i < g[x].size(); ++i) {
    if (g[x][i] != f) {
      int y = dfs(g[x][i], x);
      if (y == 1) a++;
      if (y == 2) b++;
      if (y == 3) c++;
      if (y == 4) return 4;
    }
  }
  if (a == 1 && b == 0 && c == 0) return 1;
  if (a == 2 && b == 0 && c == 0) return 2;
  if ((a > 2 || b > 0 || c > 0) && c < 2) return 3;
  if (c >= 2) {
    if (flag) {
      flag = false;
      root = x;
    }
    return 4;
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i <= n; ++i) {
    g[i].clear();
  }
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  g[1].push_back(0);
  dfs(1, 0);
  if (root == -1) root = 1;
  flag = false;
  int tot = 0;
  for (int i = 0; i < g[root].size(); ++i) {
    if (dfs(g[root][i], root) == 3) tot++;
    if (dfs(g[root][i], root) == 4) tot = 10;
  }
  if (tot > 2)
    puts("No");
  else
    puts("Yes");
  return 0;
}
