Problem:
Score : 500 points

Problem Statement

Takahashi has a string S of length N consisting of digits from 0 through 9.

He loves the prime number P. He wants to know how many non-empty (contiguous) substrings of S - there are N \times (N + 1) / 2 of them - are divisible by P when regarded as integers written in base ten.

Here substrings starting with a 0 also count, and substrings originated from different positions in S are distinguished, even if they are equal as strings or integers.

Compute this count to help Takahashi.

Constraints

1 \leq N \leq 2 \times 10^5

S consists of digits.

|S| = N

2 \leq P \leq 10000

P is a prime number.

Input

Input is given from Standard Input in the following format:

N P
S

Output

Print the number of non-empty (contiguous) substrings of S that are divisible by P when regarded as an integer written in base ten.

Sample Input 1

4 3
3543

Sample Output 1

6

Here S = 3543. There are ten non-empty (contiguous) substrings of S:

3: divisible by 3.

35: not divisible by 3.

354: divisible by 3.

3543: divisible by 3.

5: not divisible by 3.

54: divisible by 3.

543: divisible by 3.

4: not divisible by 3.

43: not divisible by 3.

3: divisible by 3.

Six of these are divisible by 3, so print 6.

Sample Input 2

4 2
2020

Sample Output 2

10

Here S = 2020. There are ten non-empty (contiguous) substrings of S, all of which are divisible by 2, so print 10.

Note that substrings beginning with a 0 also count.

Sample Input 3

20 11
33883322005544116655

Sample Output 3

68

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
#define _for(i,j,N) for(int i = (j);i < (N);i++)
#define _rep(i,j,N) for(int i = (j);i <= (N);i++)
#define _dec(i,N,j) for(int i = (N-1);i >= (j);i--)
#define _rec(i,N,j) for(int i = (N);i >=(j);i--)
#define ALL(x) x.begin(),x.end()
#define MEM(a,n) memset(a,n,sizeof(a))
#define fst first
#define scd second
#define PB push_back
#define MP make_pair

typedef long long LL;
typedef long double LD;
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

int get_int(){int x;scanf("%d",&x);return x;}
LL get_ll(){LL x;scanf("%lld",&x);return x;}
double get_db(){double x;scanf("%lf",&x);return x;}

template<typename T> int chmin(T &a,T b){if(a > b){a = b;return 1;}else return 0;}
template<typename T> int chmax(T &a,T b){if(a < b){a = b;return 1;}else return 0;}
template<typename T>
ostream& operator<<(ostream& os,const vector<T>& v) {_for(i,0,v.size()) os << v[i] << " ";return os;}
template<typename T>
ostream& operator<<(ostream& os,const set<T>& v){for(auto a:v) os << a <<" ";return os;}
template<typename T1,typename T2>
ostream& operator<<(ostream& os,const pair<T1,T2>& v){os << v.fst <<" "<<v.scd << " ";return os;}
template<typename T1,typename T2>
ostream& operator<<(ostream& os,const map<T1,T2>& v){for(auto a:v) cout << a << endl;return os;}

static const int dx[8] = { -1,  0, 1, 0,  1, -1,  1, -1};
static const int dy[8] = {0,  1, 0,  -1,  1, -1, -1,  1};
const double pi  = acos(-1.0);
const double eps = 1e-8;
const LL  INF  = 0x3f3f3f3f3f3f3f3f;
const int inf  = 0x3f3f3f3f;

const LL maxm = LL(1e18)+5;

const int maxn = 10005;

int N,P;

map<int,LL> mp;

int main()
{
    string s;
    cin >> N >> P >> s;
    int sz = s.size();

    LL ans = 0;
    LL k = 0;
    if(P == 2 || P == 5){
        int cnt = 0;
        _for(i,0,sz){
            int num = s[i] - '0';
            if(num % P == 0) cnt++;
            ans += cnt;
        }
    }
    else{
        _for(i,0,sz){
            int num = s[i] - '0';
            LL ui = (k*10 + num)%P;
            ans += mp[ui];
            mp[ui]++;
        }
    }

    cout << ans << endl;


    return 0;
}