/*
 *
 ********************************************************************************************
 * AUTHOR : Vijju123                                                                        *
 * Language: C++14                                                                          *
 * Purpose: -                                                                               *
 * IDE used: Codechef IDE.                                                                  *
 ********************************************************************************************
 *
 Comments will be included in practice problems if it helps ^^
 */



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


long long m=1000000007;
long long n,a,b,k;

int power(long long a,long long m1)
{
    if(m1==0)return 1;
    else if(m1==1)return a;
    else if(m1==2)return (1LL*a*a)%m;
    else if(m1&1)return (1LL*a*power(power(a,m1/2),2))%m;
    else return power(power(a,m1/2),2);
}
int main() {
	// your code goes here
	#ifdef JUDGE
    freopen("input.txt", "rt", stdin);
    freopen("output.txt", "wt", stdout);
    #endif
	ios_base::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>k>>a>>b;
	long long dp[10000001]={0};
	dp[1]=k%m;
	dp[2]=(1LL*k*a+1LL*b*((1LL*k*k)%m))%m;
	
	long long c1=(a+2LL*k*b)%m,c2=(1LL*a*a)%m;
	
	long long fac[10000001]={1,1},inv[10000001]={0};
	for(int i=2;i<=n;i++)
	{
	    fac[i]=(fac[i-1]*i)%m;
	}
	
	long long invFac=power(fac[n],m-2);//1/(n+1)=n!/(n+1)!. We use this for calculation of inverse in linear time.
	//This allows us to use invFactorial[i]=(i+1)*invFac[till (i+1)] (as 1/i=i!*1/(i+1)!
	
	for(int i=n;i>=1;--i)
	{
	    inv[i]=(fac[i-1]*invFac)%m;
	    invFac=(invFac*i)%m;
	}
	
	for(int i=3;i<=n;++i)
	{
	    dp[i]=(inv[i]*((m+((c1*(2*i-3))%m*dp[i-1])%m-((c2*(i-3))%m*dp[i-2])%m )%m ))%m;
	}
	
	int q,l,r;
	cin>>q;
	for(int i=1;i<=n;i++)
	    dp[i]=(dp[i-1]+(dp[i]*dp[i])%m)%m;//prefix sum. Note the 1-based indexing
	
	while(q--)
	{
	    cin>>l>>r;
	    cout<<(m+dp[r]-dp[l-1])%m<<endl;
	}
	
	
	return 0;
}