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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class D438 {

	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;
	static int[]a;
	static long[] t_sum, t_max;
	
	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n = nextInt();
		int m = nextInt();
		a = new int[n+1];
		for (int i = 1; i <= n; i++) {
			a[i] = nextInt();
		}
		t_sum = new long[4*n];
		t_max = new long[4*n];
		Build(1, 1, n);
		while (m-->0) {
			int t = nextInt();
			if (t==1) {
				int L = nextInt();
				int R = nextInt();
				pw.println(sum(1, 1, n, L, R));
			}
			else if (t==2) {
				int L = nextInt();
				int R = nextInt();
				int x = nextInt();
				assign(1, 1, n, L, R, x);
			}
			else {
				int k = nextInt();
				int x = nextInt();
				set(1, 1, n, k, x);
			}
		}
		pw.close();
	}
	private static void set(int v, int tl, int tr, int k, int x) {
		if (tl==tr) {
			t_sum[v] = x;
			t_max[v] = x;
		}
		else {
			int tm = (tl+tr) >> 1;
			if (k <= tm)
				set(2*v, tl, tm, k, x);
			else 
				set(2*v+1, tm+1, tr, k, x);
			t_sum[v] = t_sum[2*v] + t_sum[2*v+1];
			t_max[v] = Math.max(t_max[2*v], t_max[2*v+1]);
		}

	}
	private static void assign(int v, int tl, int tr, int L, int R, int x) {
		if (L > R)
			return;
		if (x > t_max[v])
			return;
		if (tl==tr) {
			t_sum[v] %= x;
			t_max[v] %= x;
		}
		else {
			int tm = (tl+tr) >> 1;
			assign(2*v, tl, tm, L, Math.min(R, tm), x);
			assign(2*v+1, tm+1, tr, Math.max(L, tm+1), R, x);
			t_sum[v] = t_sum[2*v] + t_sum[2*v+1];
			t_max[v] = Math.max(t_max[2*v], t_max[2*v+1]);
		}
		
	}
	private static long sum(int v, int tl, int tr, int L, int R) {
		if (L > R)
			return 0;
		if (tl==L && tr==R)
			return t_sum[v];
		int tm = (tl+tr) >> 1;
		return sum(2*v, tl, tm, L, Math.min(R, tm)) +
				sum(2*v+1, tm+1, tr, Math.max(L, tm+1), R);
	}
	private static void Build(int v, int tl, int tr) {
		if (tl==tr) {
			t_sum[v] = t_max[v] = a[tl]; 
		}
		else {
			int tm = (tl+tr) >> 1;
			Build(2*v, tl, tm);
			Build(2*v+1, tm+1, tr);
			t_sum[v] = t_sum[2*v] + t_sum[2*v+1];
			t_max[v] = Math.max(t_max[2*v], t_max[2*v+1]);
		}
	}
	private static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	private static long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	private static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	private static String next() throws IOException {
		while (st==null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
}