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.util.*;

public class Main {

    

    void run() throws Exception{
        int N = nextInt();
        int M = nextInt();
        int[] a = new int[N];
        for (int i = 0; i < N; i++) {
            a[i] = nextInt();
        }
        Seg seggy = new Seg(N, a);
        int l, r, x;
        for (int i = 0; i < M; i++) {
            int type = nextInt();
            switch(type){
            case 1:
                l = nextInt() - 1;
                r = nextInt() - 1;
                out.println(seggy.get(0, 0, N - 1, l, r));
                break;
            case 2:
                l = nextInt() - 1;
                r = nextInt() - 1;
                x = nextInt();
                seggy.mod(0, 0, N - 1, l, r, x);
                break;
            case 3:
                seggy.set(0, 0, N - 1  , nextInt() - 1, nextInt());
                break;
            }
        }

    }
    public static void main(String args[]) throws Exception {
            new Main().run(); // read from stdin/stdout
//              new Main("prog.in", "prog.out").run(); // or use file i/o
    }

    BufferedReader in;
    PrintStream out;
    StringTokenizer st;

    Main() throws Exception {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = System.out;
    }

    Main(String is, String os) throws Exception {
            in = new BufferedReader(new FileReader(new File(is)));
            out = new PrintStream(new File(os));
    }

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

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

}
class Seg {
        int[] arr;
        long[] sum;
        long[] max;
        int n;

        Seg(int n, int[] a) {
            this.n = n;
            arr = a;
            sum = new long[n * 4];
            max = new long[n * 4];
            build(0, 0, n - 1);
        }

        long get(int v, int l, int r, int needL, int needR) {
            if (needR < needL)
                return 0;
            if (l == needL && r == needR)
                return sum[v];
            int m = (l + r) >>> 1;
            return get(v * 2 + 1, l, m, needL, Math.min(needR, m))
                    + get(v * 2 + 2, m + 1, r, Math.max(needL, m + 1), needR);
        }

        void mod(int v, int l, int r, int needL, int needR, int mod) {
            if (needL > needR)
                return;
            if (needL == l && needR == r) {
                if (l == r) {
                    max[v] = sum[v] %= mod;
                } else {
                    if (max[v] < mod)
                        return;
                    int m = (l + r) >>> 1;
                    mod(v * 2 + 1, l, m, l, m, mod);
                    mod(v * 2 + 2, m + 1, r, m + 1, r, mod);
                    update(v);
                }
            } else {
                int m = (l + r) >>> 1;
                mod(v * 2 + 1, l, m, needL, Math.min(needR, m), mod);
                mod(v * 2 + 2, m + 1, r, Math.max(needL, m + 1), needR, mod);
                update(v);
            }
        }

        void set(int v, int l, int r, int need, int val) {
            if (l == r) {
                max[v] = sum[v] = val;
            } else {
                int m = (l + r) >>> 1;
                if (m >= need) {
                    set(v * 2 + 1, l, m, need, val);
                } else {
                    set(v * 2 + 2, m + 1, r, need, val);
                }
                update(v);
            }
        }

        void build(int v, int l, int r) {
            if (l == r) {
                sum[v] = max[v] = arr[l];
            } else {
                int m = (l + r) >>> 1;
                build(v * 2 + 1, l, m);
                build(v * 2 + 2, m + 1, r);
                update(v);
            }
        }

        void update(int v) {
            sum[v] = sum[v * 2 + 1] + sum[v * 2 + 2];
            max[v] = Math.max(max[v * 2 + 1], max[v * 2 + 2]);
        }
    }