#include <bits/stdc++.h>

using namespace std;

#define TESTC ""
#define PROBLEM "432"

#define USE_CPPIO() ios_base::sync_with_stdio(0); cin.tie(0)

#define maxn 1000005

typedef long long LL;
LL n,m;
LL x,y;
LL gia[maxn];
LL gia_prime[maxn][32];
vector <int> prime;


LL pw(LL x,LL y){

    LL ret=1,tmp=x%m;
    while(y){
        if(y&1)ret=ret*tmp%m;
        tmp=tmp*tmp%m;
        y>>=1;
    }
    return ret;
}

void build()//建階層表(mod版)
{
    prime.clear();
    LL tmp_m = m;

    for (int i = 2; i <= (LL)sqrt(tmp_m); ++i)
    {
        if (tmp_m % i)
        {
            continue;
        }
        prime.push_back(i);
        while(tmp_m % i == 0)
        {
            tmp_m /= i;
        }
    }

    if (tmp_m > 1)
    {
        prime.push_back((int)tmp_m);
    }
    gia[0] = 1;

    memset(gia_prime,0,sizeof(gia_prime));

    for (int i = 1; i < maxn; ++i)
    {
        LL tmp = i;
        for (int j = 0; j < prime.size(); ++j)
        {
            gia_prime[i][j] = gia_prime[i-1][j];
            while(tmp % prime[j] == 0)
            {
                tmp /= prime[j];
                gia_prime[i][j]++;
            }
        }
        gia[i] = gia[i-1] * tmp % m;
    }

}

void gcdex(LL a,LL b,LL c,LL *x,LL *y)//gcdex求模逆元
{
    if (b == 0)
    {
        *x = c/a;
        *y = 0;
        return;
    }

    gcdex(b,a%b,c,x,y);
    LL old_x = *x;
    *x = *y;
    *y = old_x - a/b * (*y);
    return;
}


int main()
{
    #ifdef DBG
    freopen(PROBLEM TESTC ".in", "r", stdin);
    freopen(PROBLEM ".out", "w", stdout);
    #endif

    scanf("%lld %lld",&n,&m);

    build();

    while(n--)
    {

        scanf("%lld %lld",&x,&y);
        LL ans = 1;

        for (int i = 0; i < prime.size(); ++i)
        {
            ans = ans * pw(prime[i],gia_prime[x][i] - gia_prime[x-y][i] - gia_prime[y][i]) % m;
        }

        ans = ans * gia[x] % m;

        LL xx = 0,yy = 0;
        gcdex(gia[y],m,(LL)1,&xx,&yy);
        LL tmp3 = (xx + m) % m;
        ans = ans * tmp3 % m;

        xx = 0,yy = 0;
        gcdex(gia[x-y],m,(LL)1,&xx,&yy);
        LL tmp2 = (xx + m) % m;
        ans = ans * tmp2 % m;


        printf("%lld\n",ans);
    }




    return 0;
}