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.*;
import java.math.BigInteger;
import java.util.Collections.*;
import static java.lang.Math.*;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.util.Arrays.*;
import static java.math.BigInteger.*;


public class Main{

    void run(){
        Locale.setDefault(Locale.US);
        boolean my;
        try {
            my = System.getProperty("MY_LOCAL") != null;
        } catch (Exception e) {
            my = false;
        }
        try{
            err = System.err;
            if( my ){
                sc  = new FastScanner(new BufferedReader(new FileReader("input.txt")));
//                sc  = new FastScanner(new BufferedReader(new FileReader("C:\\myTest.txt")));
                out = new PrintWriter   (new FileWriter("output.txt"));
            }
            else {
                sc = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
                out = new PrintWriter(new OutputStreamWriter(System.out));
            }
//            out = new PrintWriter(new OutputStreamWriter(System.out));
        }catch(Exception e){
            MLE();
        }
        if( my )
            tBeg = System.currentTimeMillis();
        solve();
        if( my )
            err.println( "TIME: " + (System.currentTimeMillis() - tBeg ) / 1e3 );
        exit(0);
    }

    void exit( int val ){
        err.flush();
        out.flush();
        System.exit(val);
    }

    double tBeg;
    FastScanner sc;
    PrintWriter out;
    PrintStream err;

    class FastScanner{

        StringTokenizer st;
        BufferedReader br;

        FastScanner( BufferedReader _br ){
            br = _br;
        }

        String readLine(){
            try {
                return br.readLine();
            } catch (IOException e) {
                return null;
            }
        }

        String next(){
            while( st==null || !st.hasMoreElements() )
                st = new StringTokenizer(readLine());
            return st.nextToken();
        }

        int nextInt(){ return Integer.parseInt(next()); }
        long nextLong(){ return Long.parseLong(next()); }
        double nextDouble(){ return Double.parseDouble(next()); }
    }

    void MLE(){
        int[][] arr = new int[1024*1024][]; for( int i = 0; i < 1024*1024; ++i ) arr[i] = new int[1024*1024];
    }

    void MLE( boolean doNotMLE ){ if( !doNotMLE ) MLE(); }

    void TLE(){
        for(;;);
    }

    public static void main(String[] args) {
        new Main().run();
//        new Thread( null, new Runnable() {
//            @Override
//            public void run() {
//                new Main().run();
//            }
//        }, "Lolka", 256_000_000L ).run();
    }

    ////////////////////////////////////////////////////////////////

    int n, m;
    long[] sum, max;

    void build( int v, int tl, int tr ){
        if( tl == tr ){
            max[v] = sum[v] = sc.nextInt();
        }
        else{
            int tm = (tl + tr)/2;
            build( 2*v  , tl  , tm );
            build( 2*v+1, tm+1, tr );
            sum[v] = sum[2*v] + sum[2*v+1];
            max[v] = max( max[2*v], max[2*v+1] );
        }
    }

    long getSum( int l, int r, int v, int tl, int tr ){
        if( l > r ) return 0;
        if( l == tl && r == tr ){
            return sum[v];
        }
        else{
            int tm = (tl + tr)/2;
            return
                    getSum(          l, min(r,tm), 2*v  , tl  , tm)  +
                    getSum(max(l,tm+1),         r, 2*v+1, tm+1, tr);
        }
    }

    void setMod(int l, int r, int x, int v, int tl, int tr) {
        if( l > r ) return ;

        int tm = (tl + tr)/2;

        if( l == tl && r == tr ){
            if( tl == tr ){
                sum[v] = max[v] = sum[v] % x;
            }
            else{
                if( x <= max[2*v] )
                    setMod(          l, min(r,tm), x, 2*v  , tl  , tm);
                if( x <= max[2*v+1] )
                    setMod(max(l,tm+1),         r, x, 2*v+1, tm+1, tr);
                sum[v] = sum[2*v] + sum[2*v+1];
                max[v] = max( max[2*v], max[2*v+1] );
            }
        }
        else{
            setMod(          l, min(r,tm), x, 2*v  , tl  , tm);
            setMod(max(l,tm+1),         r, x, 2*v+1, tm+1, tr);
            sum[v] = sum[2*v] + sum[2*v+1];
            max[v] = max( max[2*v], max[2*v+1] );
        }
    }

    void upd(int pos, int val, int v, int tl, int tr) {
        if( tl == tr ){
            sum[v] = max[v] = val;
        }
        else{
            int tm = (tl + tr)/2;
            if( pos <= tm )
                upd( pos, val, 2*v  , tl  , tm);
            else
                upd( pos, val, 2*v+1, tm+1, tr);
            sum[v] = sum[2*v] + sum[2*v+1];
            max[v] = max( max[2*v], max[2*v+1] );
        }
    }

    void solve(){
        n = sc.nextInt();
        m = sc.nextInt();
        sum = new long[4*(n+1)];
        max = new long[4*(n+1)];
        build(1, 1, n);
        for (int it = 0; it < m; it++) {
            int op = sc.nextInt();
            int l, r, x, k;
            if( op == 1 ){
                l = sc.nextInt();
                r = sc.nextInt();
                out.println( getSum(l,r, 1,1,n) );
            }
            else if( op == 2 ){
                l = sc.nextInt();
                r = sc.nextInt();
                x = sc.nextInt();
                setMod(l, r, x, 1, 1, n);
            }
            else if( op == 3 ){
                int pos = sc.nextInt();
                int val = sc.nextInt();
                upd( pos, val, 1,1,n );
            }
            else MLE();
        }
    }


}