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: Time Limit Exceeded)

#include<bits/stdc++.h>
#define rep(i,a,...) for(int i = (a)*(strlen(#__VA_ARGS__)!=0);i<(int)(strlen(#__VA_ARGS__)?__VA_ARGS__:(a));++i)
#define per(i,a,...) for(int i = (strlen(#__VA_ARGS__)?__VA_ARGS__:(a))-1;i>=(int)(strlen(#__VA_ARGS__)?(a):0);--i)
#define foreach(i, n) for(auto &i:(n))
#define pii pair<int, int>
#define pll pair<long long, long long>
#define all(x) (x).begin(), (x).end()
#define bit(x) (1ll << (x))
using ll = long long;
//const ll MOD = (ll)1e9+7;
const ll MOD = 998244353;
const int INF = (ll)1e9+7;
const ll INFLL = (ll)1e18;
using namespace std;
template<class t>
using vvector = vector<vector<t>>;
template<class t>
using vvvector = vector<vector<vector<t>>>;
template<class t>
using priority_queuer = priority_queue<t, vector<t>, greater<t>>;
template<class t, class u> bool chmax(t &a, u b){if(a<b){a=b;return true;}return false;}
template<class t, class u> bool chmin(t &a, u b){if(a>b){a=b;return true;}return false;}
#ifdef DEBUG
#define debug(x) cout<<"LINE "<<__LINE__<<": "<<#x<<" = "<<x<<endl;
#else
#define debug(x) (void)0
#endif

ll modpow(ll x, ll b){
    ll res = 1;
    while(b){
        if(b&1)res = res * x % MOD;
        x = x * x % MOD;
        b>>=1;
    }
    return res;
}

ll modinv(ll x){
    return modpow(x, MOD-2);
}

bool was_output = false;
template<class t>
void output(t a){
    if(was_output)cout << " ";
    cout << a;
    was_output = true;
}
void outendl(){
    was_output = false;
    cout << endl;
}
ll in(){
    ll res;
    scanf("%lld", &res);
    return res;
}
string in_s(){
    string res;
    cin >> res;
    return res;
}

template<class t>
istream& operator>>(istream&is, vector<t>&x){
    for(auto &i:x)is >> i;
    return is;
}

template<class t, class u>
istream& operator>>(istream&is, pair<t, u>&x){
    is >> x.first >> x.second;
    return is;
}

int main(){
    ll n = in();
    ll p = in();
    string str = in_s();
    ll ans = 0;
    vector<ll> sum(p,0);
    foreach(ch,str){
        int num = ch - '0';
        vector<ll> nsum(p,0);
        ++nsum[num%p];
        rep(i,p){
            nsum[(i*10+num)%p] += sum[i];
        }
        sum = nsum;
        ans += sum[0];
    }
    cout << ans << endl;
    return 0;
}