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;
vector<int> edges[100000 + 2];
bool level_two[100000 + 2];
int counts[100000 + 2];
int BFS[2][2 * 100000 + 2];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    edges[u].push_back(v);
    edges[v].push_back(u);
  }
  if (n <= 2) {
    printf("Yes\n");
    return 0;
  }
  int sz = 0;
  for (int u = 1; u <= n; u++) {
    level_two[u] = false;
    if (edges[u].size() == 1) {
      BFS[0][sz++] = u;
      level_two[u] = true;
    }
  }
  for (int iters = 1; sz > 0; iters++) {
    int nsz = 0;
    for (int j = 0; j < sz; j++) {
      int u = BFS[(iters - 1) & 1][j];
      for (int j = 0; j < edges[u].size(); j++) {
        int v = edges[u][j];
        if (!level_two[v] && edges[v].size() <= 2) {
          level_two[v] = true;
          BFS[iters & 1][nsz++] = v;
        }
      }
    }
    sz = nsz;
  }
  vector<int> new_level_two;
  for (int u = 1; u <= n; u++) {
    if (edges[u].size() != 3) continue;
    int num_level_two_neighs = 0;
    for (int j = 0; j < edges[u].size(); j++) {
      int v = edges[u][j];
      if (level_two[v]) num_level_two_neighs++;
    }
    if (num_level_two_neighs == 2) {
      new_level_two.push_back(u);
    }
  }
  for (int j = 0; j < new_level_two.size(); j++) {
    int u = new_level_two[j];
    level_two[u] = true;
  }
  bool ok = true;
  for (int u = 1; u <= n; u++) {
    if (level_two[u]) continue;
    int deg = 0;
    for (int j = 0; j < edges[u].size(); j++) {
      int v = edges[u][j];
      if (!level_two[v]) {
        deg++;
      }
    }
    if (deg > 2) {
      ok = false;
      break;
    }
  }
  if (ok) {
    printf("Yes\n");
  } else {
    printf("No\n");
  }
}
