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 long long inf = (long long)1e9 + 70;
const long long mod = 1e9 + 7;
const int maxn = 1e5 + 1000;
vector<int> order;
vector<int> adj[maxn];
vector<int> ch[maxn];
int pa[maxn];
void dfs0(int v) {
  for (int u : adj[v]) {
    if (u == pa[v]) continue;
    pa[u] = v;
    ch[v].push_back(u);
    dfs0(u);
  }
}
void dfs(int v) {
  order.push_back(v);
  for (int u : ch[v]) {
    dfs(u);
  }
}
bool oneway[maxn];
bool oneway2[maxn];
bool linear[maxn];
bool linear2[maxn];
int main() {
  if (fopen("i.txt", "r")) freopen("i.txt", "r", stdin);
  int(n);
  scanf("%d", &n);
  if (n == 1 || n == 2) {
    cout << "Yes\n";
    return 0;
  }
  for (int i = (0); i < (n - 1); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int root = -1;
  for (int i = (1); i < (n + 1); i++) {
    if (adj[i].size() == 1) continue;
    root = i;
    break;
  }
  assert(root != -1);
  dfs0(root);
  dfs(root);
  reverse((order).begin(), (order).end());
  for (int v : order) {
    oneway[v] = (ch[v].size() == 0 || (ch[v].size() == 1 && oneway[ch[v][0]]));
    linear[v] = (oneway[v] ||
                 (ch[v].size() == 2 && (oneway[ch[v][0]] && oneway[ch[v][1]])));
  }
  reverse((order).begin(), (order).end());
  for (int v : order) {
    if (v == root) continue;
    vector<bool> d;
    if (ch[pa[v]].size() >= 4) {
      oneway2[v] = false;
      linear2[v] = false;
      continue;
    }
    for (int u : ch[pa[v]]) {
      if (u != v) {
        d.push_back(oneway[u]);
      }
    }
    if (pa[v] != root) {
      d.push_back(oneway2[pa[v]]);
    }
    oneway2[v] = (d.size() == 1 && d.back());
    linear2[v] = (oneway2[v] || (d.size() == 2 && (d[0] && d[1])));
  }
  for (int v : order) {
    int ct = 0;
    if (v != root) {
      if (!linear2[v]) ct++;
    }
    for (int u : ch[v]) {
      if (!linear[u]) ct++;
    }
    if (ct > 2) {
      cout << "No\n";
      return 0;
    }
  }
  cout << "Yes\n";
  return 0;
}
