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:
/*
javac D438.java && java D438
*/

import java.io.*;
import java.util.*;

public class D438 {
	public static void main(String[] args) { new D438(); }
	FS in = new FS();
	PrintWriter out = new PrintWriter(System.out);

	int n, q;
	int[] a;
	seg tr;

	D438() {
		n = in.nextInt();
		q = in.nextInt();

		a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = in.nextInt();

		tr = new seg(0, n - 1);
		for (int i = 0; i < n; i++)
			tr.upd(i, i, a[i]);

		int t, l, r, k, x;
		while (q-- > 0) {
			t = in.nextInt();
			if (t == 1) {
				l = in.nextInt() - 1;
				r = in.nextInt() - 1;
				out.println(tr.query(l, r));
			}
			if (t == 2) {
				l = in.nextInt() - 1;
				r = in.nextInt() - 1;
				x = in.nextInt();

				tr.mod(l, r, x);
			}
			if (t == 3) {
				k = in.nextInt() - 1;
				x = in.nextInt();

				x -= tr.query(k, k);
				tr.upd(k, k, x);
			}
		}
		out.close();
	}

	class seg {
		int l, m, r;
		long sum, max, lz;
		seg L, R;
		seg (int ll, int rr) {
			l = ll;
			r = rr; 
			m = l + r >> 1;
			
			lz = 0;
			if (l == r) {
				sum = 0; 
				max = 0;
				return;
			}

			L = new seg(l, m);
			R = new seg(m + 1, r);
			sum = L.sum + R.sum;
			max = max(L.max, R.max);
		}
		void prop() {
			sum += getSum();
			max += getMax();

			L.lz += lz;
			R.lz += lz;
			lz = 0;
		}
		void fix() {
			sum = L.getSum() + R.getSum();
			max = max(L.getMax(), R.getMax());
		}
		long getSum() { return sum + lz * (r - l + 1); }
		long getMax() { return max + lz; }
		void upd(int ll, int rr, long x) {
			if (rr < l || r < ll) return;
			if (ll <= l && r <= rr) {
				lz += x; return;
			}
			prop();
			L.upd(ll, rr, x);
			R.upd(ll, rr, x);
			fix();
		}
		void mod(int ll, int rr, long x) {
			if (rr < l || r < ll) return;
			if (l == r) {
				sum = getSum() % x;
				max = getMax() % x;
				lz = 0;
				return;
			}
			prop();
			if (L.getMax() >= x) 
				L.mod(ll, rr, x);
			if (R.getMax() >= x)
				R.mod(ll, rr, x);
			fix();
		}
		long query(int ll, int rr) {
			if (rr < l || r < ll) return 0;
			if (ll <= l && r <= rr) return getSum();
			prop();
			long lq = L.query(ll, rr);
			long rq = R.query(ll, rr);
			fix();
			return lq + rq;
		}

		void debug() {
			if (l == r) {
				out.print(max + " ");
				return;
			}
			L.debug();
			R.debug();
		}
	}
	
	int abs(int x) { return x < 0 ? -x : x; }
	long abs(long x) { return x < 0 ? -x : x; }
	double abs(double x) { return x < 0 ? -x : x; }
	int min(int x, int y) { return x < y ? x : y; }	
	int max(int x, int y) { return x > y ? x : y; }
	long min(long x, long y) { return x < y ? x : y; }	
	long max(long x, long y) { return x > y ? x : y; }
	double min(double x, double y) { return x < y ? x : y; }
	double max(double x, double y) { return x > y ? x : y; }
	int gcd(int x, int y) { while (y > 0) { x = y^(x^(y = x)); y %= x; } return x; }
	long gcd(long x, long y) { while (y > 0) { x = y^(x^(y = x)); y %= x; } return x; }
	long lcm(int x, int y) { return ((long) x / gcd(x, y)) * y; }
	long lcm(long x, long y) { return (x / gcd(x, y)) * y; }

	void sort(int[] arr) {
		int sz = arr.length, j;
		Random r = new Random();
		for (int i = 0; i < sz; i++) {
			j = r.nextInt(i + 1);
			arr[i] = arr[j]^(arr[i]^(arr[j] = arr[i]));
		} Arrays.sort(arr);
	}

	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch (Exception e) {}
			} return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next()); }
		long nextLong() { return Long.parseLong(next()); }
		double nextDouble() { return Double.parseDouble(next()); }
	}
}