Problem: At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:

  1. Print operation l, r. Picks should write down the value of <image>. 
  2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r). 
  3. Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x). 



Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — initial value of array elements.

Each of the next m lines begins with a number type <image>. 

  * If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1. 
  * If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109), which correspond the operation 2. 
  * If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3. 

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Examples

Input

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3


Output

8
5


Input

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


Output

49
15
23
1
9

Note

Consider the first testcase:

  * At first, a = {1, 2, 3, 4, 5}. 
  * After operation 1, a = {1, 2, 3, 0, 1}. 
  * After operation 2, a = {1, 2, 5, 0, 1}. 
  * At operation 3, 2 + 5 + 0 + 1 = 8. 
  * After operation 4, a = {1, 2, 2, 0, 1}. 
  * At operation 5, 1 + 2 + 2 = 5. 

Solution (in JAVA):

Solution:
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;

public class Main {
    public static void mod(long[] a, long[] max, int p, int l, int r, int tl, int tr, int op) {//System.out.println(l+" "+r);
        if (max[p] < op) {//System.out.println(max[p]);
            return;
        } else if (tr < l || tl > r) {
            return;
        } else if (l == r) {
            a[p] %= op;
            max[p] = a[p];
            return;
        } else {
            mod(a, max, p * 2, l, (l + r) / 2, tl, tr, op);
            mod(a, max, p * 2 + 1, (l + r) / 2 + 1, r, tl, tr, op);
            a[p] = a[p * 2] + a[p * 2 + 1];
            if (max[p * 2] > max[p * 2 + 1]) {
                max[p] = max[p * 2];
            } else {
                max[p] = max[p * 2 + 1];
            }
        }
    }

    public static long add(long[] a, int p, int l, int r, int tl, int tr) {
        if (l >= tl && tr >= r) return a[p];
        long rzt = 0;
        if (tr < l || tl > r) return 0;//System.out.println(l + " " + r);
        rzt = add(a, p * 2, l, (l + r) / 2, tl, tr) + add(a, p * 2 + 1, (l + r) / 2 + 1, r, tl, tr);
        return rzt;
    }

    public static void change(long[] a, long[] max, int p, int l,int r, int tl, int op) {
        if (tl < l || tl > r) {
            return;
        } else if (l == r) {
            a[p] = op;
            max[p] = a[p];
            return;
        } else {
            change(a, max, p * 2, l, (l + r) / 2, tl,  op);
            change(a, max, p * 2 + 1, (l + r) / 2 + 1, r, tl, op);
            a[p] = a[p * 2] + a[p * 2 + 1];
            if (max[p * 2] > max[p * 2 + 1]) {
                max[p] = max[p * 2];
            } else {
                max[p] = max[p * 2 + 1];
            }
        }
    }

    public static void build(long[] a, long[] max, int p, int l, int r, Scanner in) {
        if (l == r) {
            a[p] = in.nextLong();
            max[p] = a[p];
        } else {
            build(a, max, p * 2, l, (l + r) / 2, in);
            build(a, max, p * 2 + 1, (l + r) / 2 + 1, r, in);
            a[p] = a[p * 2] + a[p * 2 + 1];
            if (max[p * 2] > max[p * 2+1]) {
                max[p] = max[p * 2];
            } else {
                max[p] = max[p * 2 + 1];
            }//System.out.println(max[p]);
        }
        return;
    }

    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        int n, m, op, l, r;
        long t, b = 0, k, ph, rzt = 0;
        long[] tree = new long[400050], max = new long[400050];
        n = in.nextInt();
        m = in.nextInt();
        build(tree, max, 1, 1, n, in);
        while (m-- != 0) {
            op = in.nextInt();
            l = in.nextInt();
            r = in.nextInt();
            if (op == 1) {
                System.out.println(add(tree, 1, 1, n, l, r));
            } else if (op == 2) {
                op = in.nextInt();
                mod(tree, max, 1, 1, n, l, r, op);
            } else {
                change(tree,max,1,1,n,l,r);
            }
        }
    }
}