#include <bits/stdc++.h>
using namespace std;
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define mp(a,b) make_pair((a),(b))
#define pb(a) push_back((a))
#define all(x) (x).begin(),(x).end()
#define uniq(x) sort(all(x)),(x).erase(unique(all(x)),end(x))
#define fi first
#define se second
#define dbg(x) cout<<#x" = "<<((x))<<endl
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

#define INF 2147483600

class UnionFind {
public:
  vector<int> par, rank; // parent(negative := its root and abs-value is its size), depth
  UnionFind(int sz) : par(sz, -1), rank(sz, 0){}
  int find(int x){
    if(par[x]<0) return x;
    else return par[x] = find(par[x]);
  }
  void unite(int x, int y){
    x=find(x); y=find(y);
    if(x==y) return;  // already belong to same tree
    if(rank[x] < rank[y]){  // y becomes parent node
      par[y] += par[x];
      par[x] = y;
    } else {  // x becomes parent node
      par[x] += par[y];
      par[y] = x;
      if(rank[x]==rank[y]) rank[x]++;
    }
  }
  bool same(int x, int y){ return find(x) == find(y); }
  int size(int x){ return -par[find(x)]; }
};

long calc(int m, vector<long> &vec, int x){
  UnionFind uf(m*x);
  vector<bool> c(m*x, true);
  rep(i,x) for(auto j : vec) c[m*i+j]=false;

  rep(i,m*x) if(c[i]){
    if(i%7>0 && c[i-1]) uf.unite(i-1, i);
    if(i>6 && c[i-7]) uf.unite(i-7, i);
    if(i%7<6 && c[i+1]) uf.unite(i+1, i);
    if(i+7<m*x && c[i+7]) uf.unite(i+7, i);
  }

  long ret=0;
  rep(i,m*x) if(c[i] && uf.par[i]<0) ret++;
  return ret;
}

int main(){
  long n,m,q;
  cin>>n>>m>>q;
  vector<long> a(q);
  rep(i,q) cin>>a[i];

  if(m%7==0){
    long r1 = calc(m, a, 1);
    long r2 = calc(m, a ,2);
    cout << r1 + (r2-r1)*(n/(m/7) - 1)<< endl;
  } else {
    long r1 = calc(m, a, 7);
    long r2 = calc(m, a, 14);
    cout << r1 + (r2-r1)*(n/m - 1) << endl;
  }

  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

Input Format

The input format is following:



n m q
a_1 a_2 ... a_q


Output Format

Print the number of connected part in one line.


Constraints

* n ≤ 10^{12}
* 7n is divisible by m.
* 1 ≤ q ≤ m ≤ 10^5
* 0 ≤ a_1 < a_2 < ... < a_q < m



Scoring

Subtask 1 [100 points]


* n ≤ 100000.

Subtask 2 [90 points]


* m is divisible by 7.
* a_{i + 1} - a_i = 1.

Subtask 3 [200 points]


* m is divisible by 7.

Subtask 4 [110 points]


* There are no additional constraints.



Sample Input 1


7 7 3
1 3 5


Sample Output 1


4


The calendar looks like this:


<image>

Sample Input 2


10 14 8
5 6 7 8 9 10 11 12


Sample Output 2


10


The calendar looks like this:


<image>

Output Format

Print the number of connected part in one line.


Constraints

* n ≤ 10^{12}
* 7n is divisible by m.
* 1 ≤ q ≤ m ≤ 10^5
* 0 ≤ a_1 < a_2 < ... < a_q < m



Scoring

Subtask 1 [100 points]


* n ≤ 100000.

Subtask 2 [90 points]


* m is divisible by 7.
* a_{i + 1} - a_i = 1.

Subtask 3 [200 points]


* m is divisible by 7.

Subtask 4 [110 points]


* There are no additional constraints.

Input Format

The input format is following:



n m q
a_1 a_2 ... a_q

Examples

Input

7 7 3
1 3 5


Output

4


Input

10 14 8
5 6 7 8 9 10 11 12


Output

10

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define mp(a,b) make_pair((a),(b))
#define pb(a) push_back((a))
#define all(x) (x).begin(),(x).end()
#define uniq(x) sort(all(x)),(x).erase(unique(all(x)),end(x))
#define fi first
#define se second
#define dbg(x) cout<<#x" = "<<((x))<<endl
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

#define INF 2147483600

class UnionFind {
public:
  vector<int> par, rank; // parent(negative := its root and abs-value is its size), depth
  UnionFind(int sz) : par(sz, -1), rank(sz, 0){}
  int find(int x){
    if(par[x]<0) return x;
    else return par[x] = find(par[x]);
  }
  void unite(int x, int y){
    x=find(x); y=find(y);
    if(x==y) return;  // already belong to same tree
    if(rank[x] < rank[y]){  // y becomes parent node
      par[y] += par[x];
      par[x] = y;
    } else {  // x becomes parent node
      par[x] += par[y];
      par[y] = x;
      if(rank[x]==rank[y]) rank[x]++;
    }
  }
  bool same(int x, int y){ return find(x) == find(y); }
  int size(int x){ return -par[find(x)]; }
};

long calc(int m, vector<long> &vec, int x){
  UnionFind uf(m*x);
  vector<bool> c(m*x, true);
  rep(i,x) for(auto j : vec) c[m*i+j]=false;

  rep(i,m*x) if(c[i]){
    if(i%7>0 && c[i-1]) uf.unite(i-1, i);
    if(i>6 && c[i-7]) uf.unite(i-7, i);
    if(i%7<6 && c[i+1]) uf.unite(i+1, i);
    if(i+7<m*x && c[i+7]) uf.unite(i+7, i);
  }

  long ret=0;
  rep(i,m*x) if(c[i] && uf.par[i]<0) ret++;
  return ret;
}

int main(){
  long n,m,q;
  cin>>n>>m>>q;
  vector<long> a(q);
  rep(i,q) cin>>a[i];

  if(m%7==0){
    long r1 = calc(m, a, 1);
    long r2 = calc(m, a ,2);
    cout << r1 + (r2-r1)*(n/(m/7) - 1)<< endl;
  } else {
    long r1 = calc(m, a, 7);
    long r2 = calc(m, a, 14);
    cout << r1 + (r2-r1)*(n/m - 1) << endl;
  }

  return 0;
}

```