#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define yu (998244353)
#define bas (233)
#define N 10001000
ll m,l,r,x;string s;ll ji[N];
ll k,n;
inline ll gt(ll x){return x*(x+1)/2;}
inline ll gt1(ll x){return x*(k-1)+gt(x/2)+gt((x+1)/2);}
inline ll gt2(ll x,ll y){if(!y)return 0;ll tem=(x/y),res=x%y;return x*(2*k-2)+gt(tem)*y+res*(tem+1);}
inline ll suan(ll a,ll b,ll nk){
    k=nk;if(!k)return n*(n+1)/2;
    ll an=n*(n+1)/2-k*(k-1)/2-(k-1)*(a-k+1);//cout<<a<<' '<<b<<' '<<k<<' '<<an<<'\n';
    ll z=a/k+1,g=min(b,2*k-2);ll tt=b-g,tem=(tt/z),res=tt%z;
    ll u=g+tem*2+min(2ll,res),v=b-u;ll pre=an;
    an-=gt1(u);an-=gt2(v,z-2);
    // cout<<a<<' '<<b<<' '<<k<<' '<<an<<'\n';
    return an;
}
int main(){
    // freopen("test1.in","r",stdin);
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    cin>>m>>l>>r>>x>>s;ll b=x*m;for(int i=0;i<m;i++)ji[i]=s[i]-'0',b+=ji[i];n=b;
    ll ans=0,tem=1,a=0;for(int i=0;i<l;i++)a+=x+ji[i],b-=x+ji[i],tem=tem*bas%yu;
    for(int i=l;i<=r;i++){
        ans^=suan(a,b,i)*tem%yu;
        a+=x+ji[i],b-=x+ji[i];tem=tem*bas%yu;
    }cout<<ans;return 0;
}