#include <bits/stdc++.h>
#define st first
#define en second
#define N 100034
using namespace std;

typedef long long ll;
typedef pair <int, int> pr;

struct opr{
	int id, k, b;
	opr (int id0 = 0, int k0 = 0, int b0 = 0): id(id0), k(k0), b(b0) {};
	bool operator < (const opr &b) const {return id < b.id;}
};

int cx, n, mod, q;
int a[N];
int q0, i, j;
int ch, lj, rj, k, b;
int cnt, tim, ans;
pr rg[N << 2]; // segment_tree of timestamp
opr op[5122020];

int add(int id, int L, int R){
	if(L == R){ // set st's value
		rg[id].st = cnt;
		op[cnt++] = opr(0, 1, 0);
		op[cnt++] = opr(lj, k, b);
		op[cnt++] = opr(rj + 1, 1, 0);
		if(rj < n) op[cnt++] = opr(n + 1, 0, 0);
		rg[id].en = cnt;
		return 1;
	}
	int M = L + R - 1 >> 1, i, j, K, B, pos;
	pr lr, rr;
	tim <= M ? add(id << 1, L, M) : add(id << 1 | 1, M + 1, R); // recursion
	if(tim == R){ // st union
		rg[id].st = cnt;
		lr = rg[id << 1];
		rr = rg[id << 1 | 1];
		for(pos = 0, i = lr.st + 1, j = rr.st + 1; i < lr.en || j < rr.en; ){
			K = (ll)op[i - 1].k * (ll)op[j - 1].k % mod;
			B = ((ll)op[i - 1].b * (ll)op[j - 1].k + (ll)op[j - 1].b) % mod;
			op[cnt++] = opr(pos, K, B);
			if(i < lr.en && j < rr.en && op[i].id == op[j].id){
				pos = op[i].id;
				++i; ++j;
			}else if(j >= rr.en || (i < lr.en && op[i].id < op[j].id))
				pos = op[i++].id;
			else
				pos = op[j++].id;
		}
		if(op[cnt - 1].id < n + 1) op[cnt++] = opr(n + 1, 0, 0);
		rg[id].en = cnt;
	}
	return 0;
}

int range(int id, int L, int R){
	if(lj <= L && R <= rj){ // calculate
		opr o(j, 0, 0), *oo = upper_bound(op + rg[id].st, op + rg[id].en, o) - 1;
		ans = ((ll)ans * oo->k + oo->b) % mod;
		return 1;
	}
	int M = L + R - 1 >> 1;
	if(lj <= M) range(id << 1, L, M); // recursion
	if(rj > M) range(id << 1 | 1, M + 1, R);
}

int main(){
	scanf("%d%d%d", &cx, &n, &mod);
	cx = -(cx & 1);
	for(i = 1; i <= n; i++)
		scanf("%d", a + i);
	scanf("%d", &q);
	q0 = min(q, 100000);
	cnt = tim = ans = 0;
	for(i = 0; i < q; i++){
		scanf("%d%d%d", &ch, &lj, &rj);
		if(cx){lj ^= ans; rj ^= ans;}
		if(ch == 1){
			++tim;
			scanf("%d%d", &k, &b);
			add(1, 1, q0);
		}else{
			scanf("%d", &j);
			j ^= (cx & ans);
			ans = a[j];
			//printf("a[%d] = %d\n", j, ans);
			range(1, 1, q0);
			printf("%d\n", ans);
		}
	}
	return 0;
}