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 = 1e5 + 10;
int n;
vector<int> g[N];
vector<int> ord;
int cnt[N][4], fa[N];
void dfs(int u) {
  ord.push_back(u);
  int d = min(4, (int)g[u].size()) - 1;
  cnt[u][d] = 1;
  for (int i = 0; i < (int)g[u].size(); ++i) {
    int v = g[u][i];
    if (v == fa[u]) continue;
    fa[v] = u;
    dfs(v);
    for (int j = 0; j < 4; ++j) {
      cnt[u][j] += cnt[v][j];
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; ++i) {
    scanf("%d%d", &u, &v);
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  for (int i = 0; i < n; ++i) {
    int u = ord[i], c = 0;
    for (int j = 0; j < (int)g[u].size(); ++j) {
      int v = g[u][j];
      if (v == fa[u]) {
        int tmp[4] = {cnt[0][0] - cnt[u][0], cnt[0][1] - cnt[u][1],
                      cnt[0][2] - cnt[u][2], cnt[0][3] - cnt[u][3]};
        if (g[v].size() == 3) {
          --tmp[2];
        }
        c += tmp[2] || tmp[3];
      } else {
        if (!(cnt[v][2] + cnt[v][3] == 0 ||
              (cnt[v][2] == 1 && cnt[v][3] == 0 && g[v].size() == 3))) {
          ++c;
        }
      }
    }
    if (g[u].size() >= 3 && c > 2) {
      puts("No");
      return 0;
    }
  }
  puts("Yes");
  return 0;
}
