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.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		FastScanner in = new FastScanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskD {
    public void solve(int testNumber, FastScanner in, PrintWriter out) {
		int n = in.nextInt();
		int m = in.nextInt();
		int treeSize = 1;
		while (treeSize < n) {
			treeSize *= 2;
		}
		treeSize *= 2;
		sum = new long[treeSize];
		max = new int[treeSize];
		for (int i = 0; i < n; i++) {
			int x = in.nextInt();
			modify(0, 0, n - 1, i, x);
		}
		for (int i = 0; i < m; i++) {
			int type = in.nextInt();
			if (type == 1) {
				int l = in.nextInt() - 1;
				int r = in.nextInt() - 1;
				out.println(getSum(0, 0, n - 1, l, r));
			}
			if (type == 2) {
				int l = in.nextInt() - 1;
				int r = in.nextInt() - 1;
				int mod = in.nextInt();
				divideAll(0, 0, n - 1, l, r, mod);
			}
			if (type == 3) {
				int pos = in.nextInt() - 1;
				int x = in.nextInt();
				modify(0, 0, n - 1, pos, x);
			}
		}
    }

	long[] sum;
	int[] max;

	private void divideAll(int root, int l, int r, int L, int R, int mod) {
		if (l > r || l > R || L > r || L > R) {
			return;
		}
		if (max[root] < mod) {
			return;
		}
		if (l == r) {
			sum[root] %= mod;
			max[root] = (int)sum[root];
			return;
		}
		int m = (l + r) / 2;
		divideAll(2 * root + 1, l, m, L, Math.min(R, m), mod);
		divideAll(2 * root + 2, m + 1, r, Math.max(L, m + 1), R, mod);
		sum[root] = sum[2 * root + 1] + sum[2 * root + 2];
		max[root] = Math.max(max[2 * root + 1], max[2 * root + 2]);
	}

	private long getSum(int root, int l, int r, int L, int R) {
		if (l > r || l > R || L > r || L > R) {
			return 0;
		}
		if (l == L && r == R) {
			return sum[root];
		}
		int m = (l + r) / 2;
		long res = 0;
		res += getSum(2*root + 1, l, m, L, Math.min(R, m));
		res += getSum(2*root + 2, m + 1, r, Math.max(L, m + 1), R);
		return res;
	}

	private void modify(int root, int l, int r, int pos, int val) {
		if (l == r) {
			sum[root] = val;
			max[root] = val;
			return;
		}
		int m = (l + r) / 2;
		if (pos <= m) {
			modify(2*root + 1, l, m, pos, val);
		} else {
			modify(2*root + 2, m + 1, r, pos, val);
		}
		sum[root] = sum[2*root + 1] + sum[2*root + 2];
		max[root] = Math.max(max[2*root + 1], max[2*root + 2]);
	}

}

class FastScanner {

	private BufferedReader in;
	private StringTokenizer st;

	public FastScanner(InputStream stream) {
		in = new BufferedReader(new InputStreamReader(stream));
	}

	public String next() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(in.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return st.nextToken();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

}