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;
vector<int> G[N];
int st[N];
int pre_dfs(int u, int fa) {
  int mx = 0, son = 0;
  for (auto v : G[u]) {
    if (v == fa) continue;
    mx = max(mx, pre_dfs(v, u));
    ++son;
  }
  if (mx <= 1 && son <= 1) {
    st[u] = 1;
  } else if (mx <= 1 && son <= 2) {
    st[u] = 2;
  }
  return max(mx, son);
}
int rt = -1;
int dfs(int u, int fa, int ct) {
  int c[3] = {0, 0, 0};
  for (auto v : G[u]) {
    if (v == fa) continue;
    ++c[st[v]];
  }
  if (c[0] > ct) return 0;
  for (auto v : G[u]) {
    if (v == fa || st[v]) continue;
    int nt = max(1, ct - 1 + (c[0] == 1 && c[2] == 0 && c[1] <= 2));
    if (u != rt) nt = 1;
    if (!dfs(v, u, nt)) return 0;
  }
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int u, v, i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i) {
    if ((int)G[i].size() > 2) {
      rt = i;
      break;
    }
  }
  if (rt == -1) {
    puts("Yes");
    return 0;
  }
  pre_dfs(rt, rt);
  puts(dfs(rt, rt, 2) ? "Yes" : "No");
  return 0;
}
