Problem:
Score : 600 points

Problem Statement

There is an arithmetic progression with L terms: s_0, s_1, s_2, ... , s_{L-1}.

The initial term is A, and the common difference is B. That is, s_i = A + B \times i holds.

Consider the integer obtained by concatenating the terms written in base ten without leading zeros. For example, the sequence 3, 7, 11, 15, 19 would be concatenated into 37111519. What is the remainder when that integer is divided by M?

Constraints

All values in input are integers.

1 \leq L, A, B < 10^{18}

2 \leq M \leq 10^9

All terms in the arithmetic progression are less than 10^{18}.

Input

Input is given from Standard Input in the following format:

L A B M

Output

Print the remainder when the integer obtained by concatenating the terms is divided by M.

Sample Input 1

5 3 4 10007

Sample Output 1

5563

Our arithmetic progression is 3, 7, 11, 15, 19, so the answer is 37111519 mod 10007, that is, 5563.

Sample Input 2

4 8 1 1000000

Sample Output 2

891011

Sample Input 3

107 10000000000007 1000000000000007 998244353

Sample Output 3

39122908

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

// template version 1.12
using namespace std;
#include <iostream>
#include <bits/stdc++.h>

// varibable settings
#define infile "../test/sample-1.in"
#define int long long //{{{
const int INF=1e18;
const int MOD=1e9+7; //}}}

// define basic macro {{{
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define _rrep(i,n) rrepi(i,0,n)
#define rrepi(i,a,b) for(int i=(int)((b)-1);i>=(int)(a);--i)
#define rrep(...) _overload3(__VA_ARGS__,rrepi,_rrep,)(__VA_ARGS__)
#define each(i,a) for (auto&& i : a)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define uni(x) sort(all(x));x.erase(unique(all(x)),x.end())
#define ub upper_bound
#define lb lower_bound
#define posl(A, x) (lower_bound(all(A), x)-A.begin())
#define posu(A, x) (upper_bound(all(A),x)-A.begin())
template<class T> inline void chmax(T &a, const T &b) { if((a) < (b)) (a) = (b); }
template<class T> inline void chmin(T &a, const T &b) { if((a) > (b)) (a) = (b); }

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef long double ld;
typedef pair<int,int> pii;
typedef tuple<int,int,int> iii;

template<typename T> using PQ = priority_queue<T, vector<T>, greater<T>>;
struct Fast { Fast(){ std::cin.tie(0); ios::sync_with_stdio(false); } } fast;

#if defined(PCM) || defined(LOCAL)
  #include "lib/dump.hpp"
#else
  #define dump(...) 42
  #define dump_1d(...) 42
  #define dump_2d(...) 42
#endif
//}}}

int L,A,B,M;

int digit(int x){
  int res = 0;
  int mul = 1;
  while(1){
    if (x<mul) break;
    mul *= 10;
    res += 1;
  }
  return res;
}

int bs(int n){
  int left = 0;
  int right = L+1;
  auto isright = [&](int k) {
    dump(k, digit(A+(k-1)*B));
    if (digit(A+(k-1)*B)>n) return true;
    else return false;
  };
  while (right-left>1){
    int mid = (right+left)/2;
    if (isright(mid)) { right = mid;}
    else{ left = mid; }
  }
  return right;
}

int solve(){
  cin>>L>>A>>B>>M;
  dump(L, A, B, M);

  int i = 1;
  int ans = 0;
  int mul=1;

  auto f=[&](int m){
    return A+(m-1)*B;
  };

  while(1){
    int n = digit(f(i));

    int last = bs(n)-1;
    rep(i, n*(last-i+1)) {
      ans *= 10;
      ans %= MOD;
      //TODO: speed up
    }
    ans += (f(i)+f(i+last-1))/2;
    ans %= MOD;

    if (last==L){
      break;
    }
  }

  return 0;
}

signed main() { //{{{
#ifdef INPUT_FROM_FILE
  std::ifstream in(infile);
  std::cin.rdbuf(in.rdbuf());
#endif
  solve();
  return 0;
} //}}}