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>
 
#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)
 
#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=(int)(a)-1;i>=(int)(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)
 
#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define bit(n) (1LL<<(n))
 
// #define DEBUG
 
#ifdef DEBUG
    #define dump(...) fprintf(stderr, __VA_ARGS__)
#else
    #define dump(...)
#endif
 
template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}
 
using namespace std;
using ll=long long;
using vi=vector<int>;
using vll=vector<ll>;
 
const double EPS = 1e-10;
const double PI = acos(-1.0);
const int inf = INT_MAX;
const ll INF = LLONG_MAX;
const ll MOD = 1000000007LL;
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
 
ll extgcd(ll a,ll b,ll& x,ll& y){x=1,y=0;ll g=a;if(b!=0) g=extgcd(b,a%b,y,x),y-=a/b*x;return g;}
ll ADD(const ll &a, const ll &b, const ll mod=MOD) { return (a+b)%mod;}
ll SUB(const ll &a, const ll &b, const ll mod=MOD) { return (a-b+mod)%mod;}
ll MUL(const ll &a, const ll &b, const ll mod=MOD) { return (1LL*a*b)%mod;}
ll DIV(const ll &a, const ll &b, const ll mod=MOD) {ll x,y; extgcd(b,mod,x,y);return MUL(a,(x+mod)%mod,mod);}
 
int main(void){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, P; cin >> n >> P;
    string in; cin >> in;

    if(P == 2 or P == 5){
        ll res = 0LL;
        rep(i, n){
            int x = in[i] - '0';
            if(x % P == 0){
                res += i + 1;
            }
        }
        cout << res << endl;
        return 0;
    }

    vi S(n + 1);
    vi hist(P);
    int cur = 0;
    rrep(i, n){
        int x = in[i] - '0';
        cur = 10 * cur + x;
        cur %= P;
        S[i] = cur;
    }
    for(auto& e : S) hist[e]++;

    ll res = 0LL;
    for(auto& e : hist){
        res += 1LL * e * (e - 1) / 2;
    }

    cout << res << endl;

    return 0;
}