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 static java.lang.Math.* ;
import java.util.*;
import java.io.*;

public class D
{
    void main() throws Exception {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = sc.nextInt() , m = sc.nextInt();
        int N = 1 ;
        while(N < n) N <<= 1 ;
        long [] a = new long [N + 1] ;
        for(int i = 1 ;i  <= n ;i++)
            a[i] = sc.nextInt();
        SegmentTree st = new SegmentTree(a) ;
        while(m-->0)
        {
            int type = sc.nextInt() ;
            if(type == 1)
                 out.println(st.query(sc.nextInt() , sc.nextInt()));
            else if(type == 2)
                st.update_range(sc.nextInt() , sc.nextInt() , sc.nextInt());
            else
                st.update_point(sc.nextInt() , sc.nextInt());
        }
        out.flush();
        out.close();
    }
    class SegmentTree {
        int N;
        long[] array, sTree , max;

        SegmentTree(long[] in) {
            array = in;
            N = in.length - 1;
            sTree = new long[N << 1];
            max = new long[N << 1];
            build(1, 1, N);
        }
        void build(int node, int b, int e) {
            if (b == e)
                sTree[node] = max[node] = array[b];
            else {
                int mid = b + e >> 1;
                build(node << 1, b, mid);
                build(node << 1 | 1, mid + 1, e);
                sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
                max[node] = max(max[node << 1] , max[node << 1 | 1]);
            }
        }
        void update_point(int index, int val) {
            array[index] = val ;
            index += N - 1;
            max[index] = sTree[index] = array[index - N + 1];
            while (index > 1) {
                index >>= 1;
                sTree[index] = sTree[index << 1] + sTree[index << 1 | 1];
                max[index] = max(max[index << 1] , max[index << 1 | 1]);

            }
        }
        void update_range(int i, int j, int val) { update_range(1, 1, N, i, j, val);}
        void update_range(int node, int b, int e, int i, int j, int val) {
            if (i > e || j < b || max[node] < val) return;
            if (b == e) {
                array[b] %= val;
                max[node] = sTree[node]  = array[b];
            } else {
                int mid = b + e >> 1;
                update_range(node << 1, b, mid, i, j, val);
                update_range(node << 1 | 1, mid + 1, e, i, j, val);
                sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
                max[node] = max(max[node << 1] , max[node << 1 | 1] );
            }

        }
        long query(int i, int j) { return query(1, 1, N, i, j); }
        long query(int node, int b, int e, int i, int j) {
            if (i > e || j < b) return 0;
            if (b >= i && e <= j) return sTree[node];
            int mid = b + e >> 1;
            return query(node << 1, b, mid, i, j) + query(node << 1 | 1, mid + 1, e, i, j);

        }
    }
    class Scanner
    {
        BufferedReader br;
        StringTokenizer st;

        Scanner(InputStream in)
        {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() throws Exception
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        int nextInt() throws Exception { return Integer.parseInt(next()); }

        long nextLong() throws Exception { return Long.parseLong(next()); }

        double nextDouble() throws Exception { return Double.parseDouble(next());}
    }

    public static void main (String [] args) throws Exception {(new D()).main();}

}