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

Solution (in CPP):

Solution:
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
#include<unordered_set>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const double EPS=1e-10;
const double INF=1e+10;
const double PI=acos(-1.0);
const int C_SIZE = 3121000;
namespace{
	long long fact[C_SIZE];
	long long finv[C_SIZE];
	long long inv[C_SIZE];
	long long Comb(int a,int b){
	 	if(a<b||b<0)return 0;
	 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
	}
	void init_C(int n){
		fact[0]=finv[0]=inv[1]=1;
		for(int i=2;i<n;i++){
			inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
		}
		for(int i=1;i<n;i++){
			fact[i]=fact[i-1]*i%mod;
			finv[i]=finv[i-1]*inv[i]%mod;
		}
	}
	long long pw(long long a,long long b){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%mod;
			a=a*a%mod;
			b/=2;
		}
		return ret;
	}
	int ABS(int a){return max(a,-a);}
	long long ABS(long long a){return max(a,-a);}
	double ABS(double a){return max(a,-a);}
	int sig(double r) { return (r < -EPS) ? -1 : (r > +EPS) ? +1 : 0; }
}
// ここから編集しろ
int UF[710000];
int FIND(int a){
	if(UF[a]<0)return a;return UF[a]=FIND(UF[a]);
}
void UNION(int a,int b){
	a=FIND(a);b=FIND(b);if(a==b)return;UF[a]+=UF[b];UF[b]=a;
}
int gcd(int a,int b){while(a){b%=a;swap(a,b);}return b;}
int lcm(int a,int b){return a/gcd(a,b)*b;}
int p[110000];
int main(){
	long long a;
	int b,c;scanf("%lld%d%d",&a,&b,&c);
	for(int i=0;i<c;i++){
		int t;scanf("%d",&t);
		p[t]=1;
	}
	for(int i=0;i<710000;i++)UF[i]=-1;
	int S=lcm(b,7);
	
	for(int i=0;i<S;i++){
		if(p[i%b])continue;
		if(i%7<6&&!p[(i+1)%b])UNION(i,i+1);
		if(i+7<S&&!p[(i+7)%b])UNION(i,i+7);
	}
	int sh=0;
	int sz=0;
	for(int i=0;i<S;i++)if(!p[i%b]&&UF[i]<0)sz++;
	set<pair<int,int > >s;
	for(int i=0;i<7;i++){
		if(p[i%b])continue;
		if(p[(S-7+i)%b])continue;
		s.insert(make_pair(FIND(i),FIND(S-7+i)));
	}
	sh=s.size();
	long long ret=sz+(long long)(sz-sh)*(a*7/S-1);
	printf("%lld\n",ret);
}