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;
void solve(int test_number);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.setf(ios::fixed);
  cout.precision(9);
  cerr.setf(ios::fixed);
  cerr.precision(3);
  int n = 1;
  for (int i = 0; i < n; i++) {
    solve(i);
  }
}
const int MAX_N = 300005;
int n, n_edge, deg[MAX_N];
int point[MAX_N], head[MAX_N], next_edge[MAX_N];
bool line[MAX_N];
void add_edge(int u, int v) {
  point[n_edge] = v, next_edge[n_edge] = head[u], head[u] = n_edge++;
  point[n_edge] = u, next_edge[n_edge] = head[v], head[v] = n_edge++;
}
bool is_line(int u, int p) {
  int cnt = 0;
  for (int i = head[u]; i >= 0; i = next_edge[i]) {
    int v = point[i];
    if (v == p) {
      continue;
    }
    if (!line[v]) {
      return false;
    }
    cnt++;
    if (cnt > 2) {
      return false;
    }
  }
  return true;
}
bool good(int u, int p, int lim) {
  int cnt = 0, all = 0;
  for (int i = head[u]; i >= 0; i = next_edge[i]) {
    int v = point[i];
    if (v == p) {
      continue;
    }
    if (is_line(v, u)) {
      cnt++;
    } else if (!good(v, u, 1)) {
      return false;
    }
    all++;
  }
  if (all - cnt <= lim) {
    return true;
  }
  return false;
}
void dfs_line(int u, int p) {
  int cnt = 0, lst = -1;
  for (int i = head[u]; i >= 0; i = next_edge[i]) {
    int v = point[i];
    if (v == p) {
      continue;
    }
    cnt++;
    lst = v;
    dfs_line(v, u);
  }
  if (cnt == 0 || (cnt == 1 && line[lst])) {
    line[u] = true;
  }
}
void solve(int test_number) {
  cin >> n;
  memset(head, -1, sizeof(head));
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    add_edge(u, v);
    deg[u]++;
    deg[v]++;
  }
  for (int root = 0; root < min(n, 50); root++) {
    memset(line, false, sizeof(line));
    dfs_line(root, -1);
    if (good(root, -1, 2) || is_line(root, -1)) {
      cout << "Yes" << endl;
      return;
    }
  }
  cout << "No" << endl;
}
