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:
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF438D extends PrintWriter {
	CF438D() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF438D o = new CF438D(); o.main(); o.flush();
	}

	static final int K = 320;
	int[] aa, iq;
	long[] ss;
	int[][] pq_; int[] pq;
	int[] cnt_; int cnt;
	boolean less(int u, int v) { return aa[u] > aa[v]; }
	int i2(int i) {
		return (i *= 2) > cnt ? 0 : i < cnt && less(pq[i + 1], pq[i]) ? i + 1 : i;
	}
	void pq_up(int u) {
		int i, j, v;
		for (i = iq[u]; (j = i / 2) > 0 && less(u, v = pq[j]); i = j)
			pq[iq[v] = i] = v;
		pq[iq[u] = i] = u;
	}
	void pq_dn(int u) {
		int i, j, v;
		for (i = iq[u]; (j = i2(i)) > 0 && less(v = pq[j], u); i = j)
			pq[iq[v] = i] = v;
		pq[iq[u] = i] = u;
	}
	void build(int a) {
		pq = pq_[a];
		cnt = 0;
		for (int i = a * K; i < (a + 1) * K; i++)
			pq[iq[i] = ++cnt] = i;
		for (int i = cnt / 2; i > 0; i--)
			pq_dn(pq[i]);
		cnt_[a] = cnt;
		ss[a] = query(a * K, (a + 1) * K);
	}
	long query(int l, int r) {
		long s = 0;
		for (int i = l; i < r; i++)
			s += aa[i];
		return s;
	}
	void update(int l, int r, int x) {
		for (int i = l; i < r; i++)
			aa[i] %= x;
	}
	void main() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		int b = (n + K - 1) / K;
		aa = new int[b * K];
		for (int i = 0; i < n; i++)
			aa[i] = sc.nextInt();
		iq = new int[b * K];
		pq_ = new int[b][1 + K];
		cnt_ = new int[b];
		ss = new long[b];
		for (int a = 0; a < b; a++)
			build(a);
		while (m-- > 0) {
			int t = sc.nextInt();
			if (t == 1 || t == 2) {
				int l = sc.nextInt() - 1;
				int r = sc.nextInt() - 1;
				int l_ = l / K;
				int r_ = r / K;
				if (t == 1) {
					long ans = 0;
					if (l_ == r_)
						ans = query(l, r + 1);
					else {
						ans += query(l, (l_ + 1) * K);
						for (int a = l_ + 1; a < r_; a++)
							ans += ss[a];
						ans += query(r_ * K, r + 1);
					}
					println(ans);
				} else {
					int x = sc.nextInt();
					if (l_ == r_) {
						update(l, r + 1, x);
						build(l_);
					} else {
						update(l, (l_ + 1) * K, x);
						build(l_);
						for (int a = l_ + 1; a < r_; a++) {
							pq = pq_[a];
							cnt = cnt_[a];
							while (aa[pq[1]] >= x) {
								ss[a] -= aa[pq[1]] / x * x;
								aa[pq[1]] %= x;
								pq_dn(pq[1]);
							}
							cnt_[a] = cnt;
						}
						update(r_ * K, r + 1, x);
						build(r_);
					}
				}
			} else {
				int i = sc.nextInt() - 1;
				int x = sc.nextInt();
				int a = i / K;
				ss[a] += x - aa[i];
				aa[i] = x;
				pq = pq_[a];
				cnt = cnt_[a];
				pq_dn(i);
				pq_up(i);
				cnt_[a] = cnt;
			}
		}
	}
}