#include<cstdio>
#include<vector>
#include<map>
#include<bitset>
#include<atcoder/modint>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
using namespace atcoder;
using mint = modint;
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define rrep(i, n) for (int i = (int)(n)-1; i >= 0; --i)
#define rep2(i, a, b) for (int i = (int)a; i < (int)(b); ++i)
#define rrep2(i, a, b) for (int i = (int)(b)-1; i >= (int)(a); --i)
constexpr int MAX_N = 10000000;
bitset<MAX_N+1> is_composite;//prime_divisor[i]はiを割り切る素数の１つ
vector<int> primes = {2};

int main(){
    int M, N;
    scanf("%d\n%d", &M, &N);
    if(M < N){
        printf("00000000\n");
        return 0;
    }else if(M == N){
        printf("00000001\n");
        return 0;
    }
    for (int i = 4; i <= MAX_N; i += 2){
        is_composite.set(i);
    }
    int _i = 3;
    for (int twoi; _i*_i <= MAX_N; _i += 2){
        if(!is_composite[_i]){
            primes.push_back(_i);
            twoi = _i<<1;
            for (int j = _i*_i; j <= MAX_N; j += twoi){
                is_composite.set(j);
            }
        }
    }
    for (; _i <= MAX_N; _i += 2) if(!is_composite[_i]) primes.push_back(_i);
    gp_hash_table<int, int> prime_factorization;//注目している数を素因数分解したときの<素因数, 割り切る回数>
    auto solve = [&](int x, bool add){
        int __x = x;
        for(auto &i : primes){
            if(__x < i) break;
            int cnt = 0, _x = __x;
            while((_x /= i)){
                cnt += _x;
            }
            prime_factorization[i] += add ? cnt : -cnt;
        }
    };
    solve(M, true);
    solve(N, false);
    solve(M-N, false);
    mint::set_mod(100000000);
    mint answer{1};
    for(auto [num, pow] : prime_factorization) answer *= mint::raw(num).pow(pow);
    printf("%08u\n", answer.val());
    return 0;
}