#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;
typedef long long LL;

//线段树宏定义
#define ls u << 1
#define rs u << 1 | 1

const int N = 100010;

// n:N 个非负整数   p:数模 P 的值  m:操作总数
int n, p, m;
int w[N];

struct Node {
    int l, r;
    LL sum, add, mul;
} tr[N << 2];

void pushup(int u) {
    tr[u].sum = (tr[u << 1].sum + tr[u << 1 | 1].sum) % p; //更新sum和信息
}

//计算函数
void eval(Node &t, LL add, LL mul) {
    t.sum = (t.sum * mul + (t.r - t.l + 1) * add) % p;
    t.mul = t.mul * mul % p;
    t.add = (t.add * mul + add) % p;
}

void pushdown(int u) {
    if (tr[u].add || tr[u].mul != 1) {
        eval(tr[ls], tr[u].add, tr[u].mul); //处理左儿子
        eval(tr[rs], tr[u].add, tr[u].mul); //处理右儿子
        tr[u].add = 0, tr[u].mul = 1;       //清空懒标记
    }
}

void build(int u, int l, int r) {
    tr[u] = {l, r, w[r], 0, 1};
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(ls, l, mid), build(rs, mid + 1, r);
    pushup(u);
}

void modify(int u, int l, int r, int add, int mul) {
    if (tr[u].l >= l && tr[u].r <= r) {
        eval(tr[u], add, mul);
        return;
    }

    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if (l <= mid) modify(ls, l, r, add, mul);
    if (r > mid) modify(rs, l, r, add, mul);
    pushup(u);
}

int query(int u, int l, int r) {
    if (tr[u].l >= l && tr[u].r <= r) return tr[u].sum;

    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    int sum = 0;
    if (l <= mid) sum = query(ls, l, r);
    if (r > mid) sum = (sum + query(rs, l, r)) % p;
    return sum;
}

int main() {
    //加快读入
    ios::sync_with_stdio(false), cin.tie(0);

    cin >> n >> p;
    for (int i = 1; i <= n; i++) cin >> w[i];

    //构建线段树,root=1,l=1,r=n
    build(1, 1, n);

    cin >> m;
    while (m--) {
        int t, l, r, d;
        cin >> t >> l >> r;
        if (t == 1) { //乘
            cin >> d;
            modify(1, l, r, 0, d);
        } else if (t == 2) { //加
            cin >> d;
            modify(1, l, r, d, 1);
        } else //查
            printf("%d\n", query(1, l, r));
    }
    return 0;
}