Problem: Given n numbers a0, a1, ..., an-1 and q.
I want you to perform appropriate processing for q queries.
The query has the following three types of operations.


* Shift the value
Given a pair of l and r. (l <r) Circular shift the value from al to ar.

0 1 2 3 4 5 6 7 8 9

Is given the query l = 2, r = 5.

The shifted number sequence is

0 1 5 2 3 4 6 7 8 9

Will be.


* Find the minimum value
Given a pair of l and r. (l ≤ r)
Find the smallest value from al to ar.


* Update value
Given a pair of pos and val. Update the value of apos to val.



Input

The input is given in the following format.


n q
a0
a1
..
..
..
an-1
x1 y1 z1
x2 y2 z2
..
..
..
xq yq zq


If xi = 0, it represents a shift query. At this time, let l = yi and r = zi.
If xi = 1, it represents a query for the minimum value. At this time, let l = yi and r = zi.
If xi = 2, it represents a query that updates the value. At this time, pos = yi, val = zi.


Input meets the following constraints
1 ≤ n, q ≤ 200,000
0 ≤ ai ≤ 10,000,000 For queries that update values, 0 ≤ val ≤ 10,000,000

Output

When xi = 1 is given as a query, output the value of the answer on one line.

Examples

Input

10 3
0
1
2
3
4
5
6
7
8
9
1 5 7
0 2 5
1 5 7


Output

5
4


Input

10 10
289383
930886
692777
636915
747793
238335
885386
760492
516649
641421
2 7 368690
1 3 6
0 2 6
1 1 8
0 2 9
2 2 665123
1 5 9
0 2 8
2 7 961393
1 1 2


Output

238335
238335
238335
368690

Solution (in CPP):

Solution:
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <deque>
#include <complex>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <ctime>
#include <iterator>
#include <bitset>
#include <numeric>
#include <list>
#include <iomanip>
#include <cassert>
#include <array>
#include <tuple>
#include <initializer_list>
#include <unordered_set>
#include <unordered_map>
#include <forward_list>
using ll = long long;
#define rep(j,k) for (int j = 0; j < (int)(k); j++)

namespace Treap {

    static const int INF = 1e9;

    struct Node {
        int key;
        int priority;
        Node *left, *right;
        int size;
        int min;
    };

    
    using Np = Node*;
    using pair = std::pair<Np, Np>;

    Np NIL = new Node{ -1, -1, nullptr, nullptr, 0, INF };

    int size(Np n) {
        return n->size;
    }

    int min(Np n) {
        return n->min;
    }

    void update(Np n) {
        n->size = 1 + size(n->left) + size(n->right);
        n->min = std::min({ min(n->left), n->key, min(n->right) });
    }

    void assign(Np n, int k, int x) {
        if (k < size(n->left)) {
            assign(n->left, k, x);
        } else if (size(n->left) == k) {
            n->key = x;
        } else {
            assign(n->right, k - size(n->left) - 1, x);
        }
        update(n);
    }

    int min(Np n, int l, int r) {
        l = std::max(l, 0);
        r = std::min(r, size(n));
        if (l >= r) return INF;
        if (l == 0 && r == size(n)) return min(n);

        int res = INF;
        int sl = size(n->left);
        res = std::min(res, min(n->left, l, r));
        res = std::min(res, min(n->right, l - sl - 1, r - sl - 1));
        if (l <= sl && sl < r) res = std::min(res, n->key);
        return res;
    }

    Np merge(Np l, Np r) {
        if (l == NIL) return r;
        if (r == NIL) return l;
        //assert(l->key <= r->key);
        if (l->priority > r->priority) {
            l->right = merge(l->right, r);
            update(l);
            return l;
        } else {
            r->left = merge(l, r->left);
            update(r);
            return r;
        }
    }

    pair split(Np n, int k) {
        if (n == NIL) return{ NIL, NIL };
        if (k <= size(n->left)) {
            pair p = split(n->left, k);
            n->left = p.second;
            update(n);
            return{ p.first, n };
        } else {
            pair p = split(n->right, k - size(n->left) - 1);
            n->right = p.first;
            update(n);
            return{ n, p.second };
        }
    }

    Np shift(Np n, int l, int r) {
        Np a, b, c, d;
        std::tie(a, b) = split(n, l);
        std::tie(b, c) = split(b, r - l);
        std::tie(c, d) = split(c, 1);
        return merge(merge(a, c), merge(b, d));
    }

    Np append(Np n, int k) {
        static Node pool[200010];
        static int p = 0;
        Np a = pool + p++;
        *a = { k, rand(), NIL, NIL, 1, k };
        return merge(n, a);
    }
};

using namespace std;
int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);

    int n, q;
    cin >> n >> q;

    Treap::Node *root = Treap::NIL;

    for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        root = Treap::append(root, a);
    }

    //Treap::show(root);

    for (int i = 0; i < q; i++) {
        int x, y, z;
        cin >> x >> y >> z;
        if (x == 0) {
            root = Treap::shift(root, y, z);
        }
        if (x == 1) {
            //puts("=====");
            //Treap::show(root);
            cout << Treap::min(root, y, z + 1) << '\n';
        }
        if (x == 2) {
            Treap::assign(root, y, z);
        }
    }
}