#include<iostream>
#define int long long
using namespace std;
int a[1145141],f[1145141],d[1145141],f2[1145141]; 
//a表示原数组，f表示懒标记加，f2表示懒标记乘，d表示线段树
int n,q,mod;

void pushdamn(int s,int t,int p){
	int mid=(s+t)>>1;
	d[p*2]=(d[p*2]*f2[p]+f[p]*(mid-s+1))%mod;
	d[p*2+1]=(d[p*2+1]*f2[p]+f[p]*(t-mid))%mod;

	f2[p*2]=(f2[p]*f2[p*2])%mod;
	f2[p*2+1]=(f2[p]*f2[p*2+1])%mod;

	f[p*2]=(f[p*2]*f2[p]+f[p])%mod;
	f[p*2+1]=(f[p*2+1]*f2[p]+f[p])%mod;

	f2[p]=1;
	f[p]=0;
	return;
}

void build(int s,int t,int p){
	f2[p]=1;
	f[p]=0;
	if (s==t){
		d[p]=a[s];
		d[p]%=mod;
		return;
	}
	int mid=(s+t)>>1;
	build(s,mid,p*2);
	build(mid+1,t,p*2+1);
	d[p]=(d[p*2+1]+d[p*2])%mod;
	return;
}

void update(int l,int r,int s,int t,int c,int p){			//加法更新
	if (l<=s && t<=r){
		d[p]+=(t-s+1)*c;
		f[p]+=c;
		d[p]%=mod;
		f[p]%=mod;
		return;
	}
	int mid=(s+t)>>1;
	pushdamn(s,t,p);
	if (l<=mid)update(l,r,s,mid,c,p*2);
	if (mid<r)update(l,r,mid+1,t,c,p*2+1);
	d[p]=(d[p*2]+d[p*2+1])%mod;
	return;
}

void update2(int l,int r,int s,int t,int c,int p){			//乘法更新
	if (l<=s && t<=r){
		d[p]*=c;
		f2[p]*=c;
		f[p]*=c;
		d[p]%=mod;
		f2[p]%=mod;
		f[p]%=mod;
		return;
	}
	int mid=(s+t)>>1;
	pushdamn(s,t,p);
	if (l<=mid)update2(l,r,s,mid,c,p*2);//update改为 update2
	if (mid<r)update2(l,r,mid+1,t,c,p*2+1);//update改为 update2
	d[p]=(d[p*2]+d[p*2+1])%mod;
	return;
}

int getsum(int l,int r,int s,int t,int p){
	if (l<=s && t<=r)return d[p];
	pushdamn(s,t,p);
	int mid=(s+t)>>1;
	int sum=0;
	if (l<=mid)sum+=getsum(l,r,s,mid,p*2);//l<=s改为 l<=mid 
	if (r>mid)sum+=getsum(l,r,mid+1,t,p*2+1);//t<r改为 r>mid 
	return sum;
}

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> n >> q >> mod;
    for (int i=1;i<=n;i++){
        cin >> a[i];
    }
	build(1,n,1);
    while(q--){
        int ope,x,y;
        cin >> ope >> x >> y;
        if (ope==1){
            int k;
            cin >> k;
            update2(x,y,1,n,k,1);
        }else if(ope==2){
            int k;
            cin >> k;
            update(x,y,1,n,k,1);
        }else{
			cout << getsum(x,y,1,n,1)%mod << "\n";
		}
    }
	return 0;
}