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<vector<int> > adj;
vector<int> VIS;
int LVL;
vector<int> ATT;
int dfs(int u) {
  LVL++;
  VIS[u] = 1;
  vector<int> V;
  int hc = 0, oc = 0, ohc = 0, tc = 0;
  for (int i = 0; i < adj[u].size(); ++i) {
    int v = adj[u][i];
    if (VIS[v]) continue;
    int t = dfs(v);
    if (t == 0) {
      return 0;
    }
    if (t == 1) {
      hc++;
    }
    if (t == 2) {
      oc++;
    }
    if (t == 3) {
      ohc++;
    }
    if (t == 4) {
      tc++;
    }
  }
  int a = 0;
  if (tc) {
    a = 0;
  } else if (ohc) {
    if (ohc <= 1)
      a = 3;
    else if (ohc <= 2)
      a = 4;
    else
      a = 0;
  } else if (oc) {
    a = 3;
  } else {
    if (hc <= 1)
      a = 1;
    else if (hc <= 2)
      a = 2;
    else
      a = 3;
  }
  ATT[u] = a;
  LVL--;
  return a;
}
int main() {
  int N;
  cin >> N;
  adj = vector<vector<int> >(N);
  VIS = vector<int>(N);
  ATT = vector<int>(N);
  for (int i = 0; i < N - 1; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int t = dfs(0);
  for (int i = 0; i < N; ++i) {
    if (ATT[i] == 4) {
      VIS = vector<int>(N);
      t = dfs(i);
      break;
    }
  }
  if (t == 0) {
    cout << "No" << endl;
  } else {
    cout << "Yes" << endl;
  }
}
