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:
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
What do you think? What do you think?
1st on Billboard, what do you think of it
Next is a Grammy, what do you think of it
However you think, I’m sorry, but shit, I have no fcking interest
*******************************
I'm standing on top of my Monopoly board
That means I'm on top of my game and it don't stop
til my hip don't hop anymore
https://www.a2oj.com/Ladder16.html
*******************************
300iq as writer = Sad!
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class x438D
   {
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         int M = Integer.parseInt(st.nextToken());
         int[] arr = new int[N];
         st = new StringTokenizer(infile.readLine());
         for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
         SegTree segtree = new SegTree(N);
         for(int i=0; i < N; i++)
            segtree.updateVal(i, arr[i]);
         StringBuilder sb = new StringBuilder();
         while(M-->0)
         {
            st = new StringTokenizer(infile.readLine());
            int type = Integer.parseInt(st.nextToken());
            if(type == 1)
            {
               int L = Integer.parseInt(st.nextToken())-1;
               int R = Integer.parseInt(st.nextToken())-1;
               sb.append(segtree.query(L, R)+"\n");
            }
            else if(type == 2)
            {
               int L = Integer.parseInt(st.nextToken())-1;
               int R = Integer.parseInt(st.nextToken())-1;
               int mod = Integer.parseInt(st.nextToken());
               segtree.update(L, R, mod);
            }
            else
            {
               int K = Integer.parseInt(st.nextToken())-1;
               int X = Integer.parseInt(st.nextToken());
               segtree.updateVal(K, X);
            }
         }
         System.out.print(sb);
      }
   }
   class SegTree
   {
      //definitions
      private long NULL = 0L;
      //ds stuff
      private long[] tree;
      private long[] max;
      private int length;
      
      public SegTree(int N)
      {
         length = N;   int b;
         for(b=0; (1<<b) < length; b++);
         tree = new long[1<<(b+1)];
         max = new long[1<<(b+1)];
      }
      public long query(int left, int right)
      {
         //left and right are 0-indexed
         return get(1, 0, length-1, left, right);  
      }
      private long get(int v, int currL, int currR, int L, int R)
      {
         if(L > R)
            return NULL;
         if(L <= currL && currR <= R)
            return tree[v];
         int mid = (currL+currR)/2;
         return get(v*2, currL, mid, L, Math.min(R, mid))+
            get(v*2+1, mid+1, currR, Math.max(L, mid+1), R);
      }
      public void update(int left, int right, int mod)
      {
         add(1, 0, length-1, left, right, mod);
      }
      private void add(int v, int currL, int currR, int L, int R, int mod)
      {
         if(L > R || max[v] < mod)
            return;
         if(currL == currR)
         {
            //exact covering
            tree[v] %= mod;
            max[v] %= mod;
            return;
         }
         int mid = (currL+currR)/2;
         add(v*2, currL, mid, L, Math.min(R, mid), mod);
         add(v*2+1, mid+1, currR, Math.max(L, mid+1), R, mod);
         tree[v] = tree[v*2]+tree[v*2+1];
         max[v] = Math.max(max[v*2], max[v*2+1]);
      }
      public void updateVal(int index, int val)
      {
         add2(1, 0, length-1, index, index, val);
      }
      private void add2(int v, int currL, int currR, int L, int R, int val)
      {
         if(L > R)
            return;
         if(currL == currR)
         {
            //exact covering
            tree[v] = val;
            max[v] = val;
            return;
         }
         int mid = (currL+currR)/2;
         add2(v*2, currL, mid, L, Math.min(R, mid), val);
         add2(v*2+1, mid+1, currR, Math.max(L, mid+1), R, val);
         tree[v] = tree[v*2]+tree[v*2+1];
         max[v] = Math.max(max[v*2], max[v*2+1]);
      }
   }