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>
using namespace std;
#define inf INT_MAX
#define INF LLONG_MAX
#define ll long long
#define ull unsigned long long
#define M (int)(1e9+7)
#define P pair<int,int>
#define FOR(i,m,n) for(int i=(int)m;i<(int)n;i++)
#define RFOR(i,m,n) for(int i=(int)m;i>=(int)n;i--)
#define rep(i,n) FOR(i,0,n)
#define rrep(i,n) RFOR(i,n,0)
#define all(a) a.begin(),a.end()
const int vx[4] = {0,1,0,-1};
const int vy[4] = {1,0,-1,0};
#define F first
#define S second
#define PB push_back
#define EB emplace_back
#define int ll
#define vi vector<int>
#define IP pair<int,P>
#define PI pair<P,int>
#define PP pair<P,P>
#define Yes(f){cout<<(f?"Yes":"No")<<endl;}
#define YES(f){cout<<(f?"YES":"NO")<<endl;}


signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<fixed<<setprecision(20);

    int n,p;
    string s;
    cin>>n>>p>>s;
    
    int sum[2][100000]={};
    sum[0][0]=1;
    int ans=0;
    set<int> st[2];
    st[0].insert(0);

    rep(i,n){
        for(int x:st[i&1]){
            int y=x*10+s[i]-'0';
            y%=p;
            sum[(i+1)&1][y]+=sum[i&1][x];
            st[(i+1)&1].insert(y);
        }
        ans+=sum[(i+1)&1][0];
        rep(x,p){
            sum[i&1][x]=0;
        }
        st[(i+1)&1].insert(0);
        st[i&1].clear();
        sum[(i+1)&1][0]++;
    }

    cout<<ans<<endl;


}