#include "bits/stdc++.h"
using namespace std;
#define FOR(i,j,k) for(int (i)=(j);(i)<(int)(k);++(i))
#define rep(i,j) FOR(i,0,j)
#define each(x,y) for(auto &(x):(y))
#define mp make_pair
#define mt make_tuple
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<(x)<<endl
#define smax(x,y) (x)=max((x),(y))
#define smin(x,y) (x)=min((x),(y))
#define MEM(x,y) memset((x),(y),sizeof (x))
#define sz(x) (int)(x).size()
#define pb push_back
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;

class UnionFind {
    int cnt;
    vector<int> par, rank, size;
public:
    UnionFind() {}
    UnionFind(int _n) :cnt(_n), par(_n), rank(_n), size(_n, 1) {
        for (int i = 0; i<_n; ++i) par[i] = i;
    }
    int find(int k) {
        return (k == par[k]) ? k : (par[k] = find(par[k]));
    }
    int operator[](int k) {
        return find(k);
    }
    int getSize(int k) {
        return size[find(k)];
    }
    void unite(int x, int y) {
        x = find(x); y = find(y);
        if (x == y) return;
        --cnt;
        if (rank[x] < rank[y]) {
            par[x] = y;
            size[y] += size[x];
        } else {
            par[y] = x;
            size[x] += size[y];
            if (rank[y] == rank[x]) ++rank[x];
        }
    }
    int count() {
        return cnt;
    }
};

bool A[100000 * 7][7];
int f(int h, int m, vi &col) {
    MEM(A, 0);
    UnionFind uf(7 * h);
    int cur = 0, black = 0;
    rep(y, h)rep(x, 7) {
        if (!col[cur%m]) {
            A[y][x] = 1;
            if (y > 0 && A[y - 1][x])uf.unite(cur, cur - 7);
            if (x > 0 && A[y][x - 1])uf.unite(cur, cur - 1);
        } else {
            black++;
        }
        cur++;
    }
    return uf.count() - black;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n;
    int m, q;
    cin >> n >> m >> q;
    vi col(m);
    rep(i, q) {
        int a;
        cin >> a;
        col[a] = 1;
    }

    int H = m;
    if (H % 7 == 0)H /= 7;
    int b = f(H, m, col), a = f(H * 2, m, col) - b;
    cout << a*((n - H) / H) + b << endl;
}

### Prompt

Your challenge is to write 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 FOR(i,j,k) for(int (i)=(j);(i)<(int)(k);++(i))
#define rep(i,j) FOR(i,0,j)
#define each(x,y) for(auto &(x):(y))
#define mp make_pair
#define mt make_tuple
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<(x)<<endl
#define smax(x,y) (x)=max((x),(y))
#define smin(x,y) (x)=min((x),(y))
#define MEM(x,y) memset((x),(y),sizeof (x))
#define sz(x) (int)(x).size()
#define pb push_back
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;

class UnionFind {
    int cnt;
    vector<int> par, rank, size;
public:
    UnionFind() {}
    UnionFind(int _n) :cnt(_n), par(_n), rank(_n), size(_n, 1) {
        for (int i = 0; i<_n; ++i) par[i] = i;
    }
    int find(int k) {
        return (k == par[k]) ? k : (par[k] = find(par[k]));
    }
    int operator[](int k) {
        return find(k);
    }
    int getSize(int k) {
        return size[find(k)];
    }
    void unite(int x, int y) {
        x = find(x); y = find(y);
        if (x == y) return;
        --cnt;
        if (rank[x] < rank[y]) {
            par[x] = y;
            size[y] += size[x];
        } else {
            par[y] = x;
            size[x] += size[y];
            if (rank[y] == rank[x]) ++rank[x];
        }
    }
    int count() {
        return cnt;
    }
};

bool A[100000 * 7][7];
int f(int h, int m, vi &col) {
    MEM(A, 0);
    UnionFind uf(7 * h);
    int cur = 0, black = 0;
    rep(y, h)rep(x, 7) {
        if (!col[cur%m]) {
            A[y][x] = 1;
            if (y > 0 && A[y - 1][x])uf.unite(cur, cur - 7);
            if (x > 0 && A[y][x - 1])uf.unite(cur, cur - 1);
        } else {
            black++;
        }
        cur++;
    }
    return uf.count() - black;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n;
    int m, q;
    cin >> n >> m >> q;
    vi col(m);
    rep(i, q) {
        int a;
        cin >> a;
        col[a] = 1;
    }

    int H = m;
    if (H % 7 == 0)H /= 7;
    int b = f(H, m, col), a = f(H * 2, m, col) - b;
    cout << a*((n - H) / H) + b << endl;
}

```