#include<stdio.h>
#include<iostream>
#include<vector>
#include<stack>
#include<utility>//pair
#include<functional>//greater
#include<queue>
#include<set>
#include<string.h>
#include<algorithm>
#include<math.h>
#define F first
#define S second
#define sd(n) scanf("%d",&n)
#define sdd(m,n) scanf("%d %d",&m,&n)
#define sl(n) scanf("%lld",&n)
#define sll(m,n) scanf("%lld %lld",&m,&n)
#define sc(n) scanf("%c",&n)
#define ss(n) scanf("%s",n)
#define pd(n) printf("%d\n",n)
#define pl(n) printf("%lld\n",n)
#define pc(n) printf("%c ",n)
#define ps(n) printf("%s ",n)
#define NL printf("\n")
#define pb(n) push_back(n)
#define sz size()
#define clr(v,a,b) for(int i=a;i<=b;i++)v[i].clear()
#define all(v) v.begin(),v.end()
#define mod 1000000007//10^9+7
#define pp(a,b) make_pair(a,b)
/*#include<fstream>
ifstream fin("in.txt");
ofstream fout("out.txt");*/
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const int MXN=100005;
//two approach for storing more than two values-node structure and separate arrays
ll ar[100005];
ll ia[100005];
ll mb[100005];
struct bit
{
    ll a[MXN];
    bit()
    {
        memset(a,0,sizeof(a));
    }
    void upd(int i,ll x,int n)
    {
        if(i==0)
            {a[0]=(a[0]+x+mod)%mod;return;}
        while(i<=n)
        {
            a[i]=(a[i]+x+mod)%mod;
            i+=i & -i;
        }
    }
    ll sum(int i)
    {
        ll s=0;
        while(i>0)
        {
            s=(s+a[i])%mod;
            i-= i & -i;
        }
        s=(s+a[0])%mod;
        return s;
    }

};
long long fpow(long long x,long long n)
{
    long long r=1;
    while(n>0)
    {
        if(n%2)
        {
            r=(r*x)%mod;
        }
        x=(x*x)%mod;
        n/=2;
    }
    return r;
}
int main()
{
    //int a[10]={3,5,2,5,8,1,5,0,9,4};
    ll n,a,b,q;
    sl(n);sll(a,b);sl(q);
    int bb=mod-b;
    ia[0]=1;mb[0]=1;
    ia[1]=fpow(a,mod-2);
    mb[1]=bb;
    for(int i=2;i<=n+2;i++)
    {
        ia[i]=(ia[1]*ia[i-1])%mod;
        mb[i]=(bb*mb[i-1])%mod;
    }
    bit tr;
    for(int i=0;i<n;i++)
    {
        ll c;
        sl(c);
        ar[i]=(((c*mb[i])%mod)*ia[i])%mod;
        tr.upd(i,ar[i],n);
    }
    while(q--)
    {
        int t,p,q;
        sd(t);sdd(p,q);
        if(t==1)
        {
            ll nv=((((ll)q*mb[p])%mod)*ia[p])%mod;
            ll uv=(nv-ar[p]+mod)%mod;
            tr.upd(p,uv,n);
            ar[p]=nv;
        }
        else
        {
            ll ans=tr.sum(q);
            if(p>0)
                ans=(ans-tr.sum(p-1)+mod)%mod;
            ll px=(mb[p]*ia[p])%mod;
            ll iv=fpow(px,mod-2);
            ans=(ans*iv)%mod;
            if(ans==0)
                printf("Yes\n");
            else
                printf("No\n");
        }
    }
    return 0;
}