#include<algorithm>
#include<complex>
#include<ctype.h>
#include<iomanip>
#include<iostream>
#include<fstream>
#include<map>
#include<math.h>
#include<numeric>
#include<queue>
#include<set>
#include<stack>
#include<stdio.h>
#include<string>
#include<string>
#include<vector>

using namespace std;
typedef long long ll;

#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<endl
#define p2(s, t) cout << (s) << " " << (t) << endl
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << endl
#define p_yes() p("Yes")
#define p_no() p("No")

const ll mod = 1e9 + 7;
const ll inf = 1e18;

/*
ABCDEFGHIJKLMNOPQRSTUVWXYZ 26文字

足りない部分にスペースを入れてAとZの位置を揃えていくと、足りないのが見つけやすいんじゃないかな

ABCDEFGHIJKLMNOPQRSTUVWX Z  Y
ABCDEFGHIJKLMN PQRSTVWXYZ_  O
 BCDEFGHIJKLMNOPQSTUVWXYZ   U
ABCD FGHIJKLMNOPQRSTUVWXYZ_ E
ABCDEF H JKLMNOPQRSTU WXYZ   G I V
ABCD FGHIJKLM OPQRSTUVWXYZ_ E
ABCDEFGHIJKLMNOPQRS UV XYZ T W
ABCDEFGHIJKLMN PQRSTUVWXYZ_ O
ABCDEFGHJKLMOPQRSUVWXYZ
ABCDFHIJKLMNOPQRSTUVWXYZ
ABCDFGHIJKLMNOPQ TUVWXYZ_
ABCDEFGHIJKLMOPQRSTUVWXYZ_
BCDEFGHIJKLMOPQRSTUVWXYZ
ABCEFGHIJKLMNOPQRSTUVWXYZ_
ABC DEFGHIJKLNOPQRSTUVWXYZ._
ABCDEFGHIJKLMNOPQRSTUVWXZ
ABCDEFGHIJKLMNPQRSTVWXYZ
ABCDEFGHIJKLMN OPQSTUVWXYZ_
ABCDEFGHIJKLMNOPQRSUVWXYZ
BCDEFGHIJKLMNOPQRTUVWXYZ
ABCDEFGHIJLMNOPQRSTUVWXYZ_ 
ABCDEFGHJKLMNOPQRTUVWXYZ_
ABCDEFGHIJKLMNOPQRSUVWXYZ
ABCDEFGHIJKLMNPQRSTUVWXYZ_
ABCDEGHJKLM OPQRSTUVWXYZ
ABCEFGHIJKLMNOPQRSTUVWXYZ_
ABCDEFGHIJKLMNOPQRSUVWXYZ
ABCDEFGIJKLMNOPQRSTUVWXY Z
ABCDFGHIJKLMNOPQRSTUVWXYZ_
ABCDEGHIJKLMNOPQRSTUVWXYZ
BDEFGHIJKLMNOPQRSUVWXYZ
ABCDEFGHIJKL MNPQSTUVWXYZ
ABCDEFGHJKLMNOPQRSTUVWXYZ
BCDEFGHIJKMNOPQRSTUVWXYZ_
ABCDEFGHIJKLMNPQRSTUVWXYZ 
ABCDEGHIJKLMNOPQRSTUVWXYZ_
ABCDEFGHIJKLMOPQRSTUVWXYZ,_
ABCDEFGHIJKLNPQRSTUVWXYZ
ABCEFGHIJK LMNOPQRSTVWXYZ
ABCDEFGHIJKMNPQRSTUVWXYZ_
ABCDEFGHIJKLNOPQRSTUVWXYZ.
*/

ll factorial_mod(ll n, ll M){
    ll ans = 1;
    FOR(i, 1, n+1){
        ans *= i;
        ans %= M;
    }
    return ans;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    ll N, M;
    cin >> N >> M;

    // YOU_ARE_GIVEN_TWO_INTEGERS_N_AND_M._YOUR_TASK_IS_TO_FIND_THE_FACTORIAL_OF_N,_MODULO_M.

    if(N>=M){
        p(0);
        return 0;
    }

    ll ans = factorial_mod(N, M);
    p(ans);

    return 0;
}