• 直接维护乘积是肯定不可行的， 精度会爆炸， 于是我们来维护对数的和， 最后来计算最高位即可
• 那么转换成区间求和， 区间排序
• 区间排序的方式可以采用线段树维护最大递增块来解决，外层用set来维护线段树的区间， 然后利用线段树的合并分裂性质来操作即可

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#include<set>
#include<cmath>
#define ll long long
#define M (1 << 18)
#define N 20000010
#define double long double
const double eps = 1e-8;
using namespace std;
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}
double c[M], ver[M];
int num[M], n, m;
int lowbit(int x) {
    return x & -x;
}
void add(int x, double v) {
    for(int i = x; i <= n; i += lowbit(i)) c[i] += v;
}

double query(int x) {
    double ans = 0;
    for(int i = x; i; i -= lowbit(i)) ans += c[i];
    return ans;
}

struct Note {
    int l, r, rt, op;
    Note(int ln = 0, int rn = 0, int rtn = 0, int opn = 0) {
        l = ln, r = rn, rt = rtn, op = opn;
    }
    bool operator < (const Note &b) const {
        return this->l < b.l;
    }
};
#define S set<Note>::iterator
set<Note> st;
int ls[N], rs[N], cnt[N], f;
double sum[N];

void pushup(int now) {
    cnt[now] = cnt[ls[now]] + cnt[rs[now]];
    sum[now] = sum[ls[now]] + sum[rs[now]];
}

int merge(int x, int y) {
    if(!x || !y) return x + y;
    ls[x] = merge(ls[x], ls[y]);
    rs[x] = merge(rs[x], rs[y]);
    cnt[x] = cnt[x] + cnt[y];
    sum[x] = sum[x] + sum[y];
    return x;
}

S insert(Note x) {
    add(x.l, sum[x.rt]);
    return st.insert(x).first;
}

void Del(S it) {
    add(it->l, -sum[it->rt]);
    st.erase(it);
}

void split(int x, int &rt1, int &rt2, int l, int r, int k) {
    rt1 = ++f;
    rt2 = ++f;
    if(l == r) {
        cnt[rt1] = k;
        sum[rt1] = ver[l] * k;
        cnt[rt2] = cnt[x] - cnt[rt1];
        sum[rt2] = sum[x] - sum[rt1];
        return;
    }
    int mid = (l + r) >> 1;
    if(cnt[ls[x]] >= k) {
        rs[rt2] = rs[x];
        split(ls[x], ls[rt1], ls[rt2], l, mid, k);
    } else {
        ls[rt1] = ls[x];
        split(rs[x], rs[rt1], rs[rt2], mid + 1, r, k - cnt[ls[x]]);
    }
    pushup(rt1);
    pushup(rt2);
}

S split(int x) {
    if(x > n) return st.end();
    S it = st.upper_bound(Note(x, 0, 0, 0));
    it--;
    Note hh = *it;
    if(hh.l == x) return it;
    int rt1, rt2;
    if(!hh.op) split(hh.rt, rt1, rt2, 1, n, x - hh.l);
    else split(hh.rt, rt2, rt1, 1, n, hh.r - x + 1);
    Del(it);
    insert(Note(hh.l, x - 1, rt1, hh.op));
    return insert(Note(x, hh.r, rt2, hh.op));
}

void build(int &x, int l, int r, int k) {
    x = ++f;
    cnt[x]++;
    sum[x] += ver[k];
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(k <= mid) build(ls[x], l, mid, k);
    else build(rs[x], mid + 1, r, k);
}


int calc(int l, int r) {
    S L = split(l), R = split(r + 1);
    R--;
    double ans = query(R->r) - query(L->l - 1);
    double out = pow(10, ans - floorl(ans) + eps);
    return floorl(out);
}

void updata(int l, int r, int op) {
    S L = split(l);
    split(r + 1);
    int rt = 0;
    for(S it = L; it != st.end() && (it->l) <= r; Del(it++)) {
        rt = merge(rt, it->rt);
    }
    insert(Note(l, r, rt, op));
}

int main() {
    n = read(), m = read();
    for(int i = 1; i <= n; i++) num[i] = read(), ver[i] = log10(i);
    for(int i = 1; i <= n; i++) {
        int now;
        build(now, 1, n, num[i]);
        insert(Note(i, i, now, 0));
    }
    while(m--) {
        int op = read(), l = read(), r = read();
        if(op == 2) cout << calc(l, r) << "\n";
        else {
            op = read() ^ 1;
            updata(l, r, op);
        }
    }
    return 0;
}