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.*;

public class D {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;

	int ROOT;
	int K;

	void update(int part) {
		long curSum = 0;
		int curMax = -1;
		for (int i = part * ROOT; i < (part + 1) * ROOT; i++) {
			curSum += arr[i];
			curMax = Math.max(curMax, arr[i]);
		}
		max[part] = curMax;
		sum[part] = curSum;
	}

	int[] arr, max;
	long[] sum;

	void solve() throws IOException {
		int n = nextInt();
		int qq = nextInt();

		ROOT = (int) (Math.sqrt(n) + 1);
		K = n / ROOT + 1;

		arr = new int[ROOT * K];
		max = new int[K];
		sum = new long[K];

		for (int i = 0; i < n; i++) {
			arr[i] = nextInt();
		}

		for (int i = 0; i < K; i++) {
			update(i);
		}

		while (qq-- > 0) {
			int type = nextInt();
			if (type == 1) {
				int l = nextInt() - 1;
				int r = nextInt() - 1;

				int partL = l / ROOT;
				int partR = r / ROOT;
				long outp = 0;
				if (partL == partR) {
					for (int i = l; i <= r; i++) {
						outp += arr[i];
					}
				} else {

					for (int i = l; i < (partL + 1) * ROOT; i++) {
						outp += arr[i];
					}
					for (int i = partR * ROOT; i <= r; i++) {
						outp += arr[i];
					}
					for (int i = partL + 1; i < partR; i++) {
						outp += sum[i];
					}
				}
				out.println(outp);

			} else if (type == 2) {
				int l = nextInt() - 1;
				int r = nextInt() - 1;
				int x = nextInt();

				int partL = l / ROOT;
				int partR = r / ROOT;

				// System.err.println(partL + " " + partR);

				if (partL == partR) {
					if (max[partL] >= x) {
						for (int i = l; i <= r; i++) {
							arr[i] %= x;
						}
						update(partL);
					}
				} else {

					if (max[partL] >= x) {

						for (int i = l; i < (partL + 1) * ROOT; i++) {
							arr[i] %= x;
						}
						update(partL);
					}
					if (max[partR] >= x) {
						for (int i = partR * ROOT; i <= r; i++) {
							arr[i] %= x;
						}
						update(partR);
					}
					for (int i = partL + 1; i < partR; i++) {
						if (max[i] >= x) {
							for (int j = i * ROOT; j < (i + 1) * ROOT; j++) {
								arr[j] %= x;
							}
							update(i);
						}
					}
				}
			} else if (type == 3) {
				int k = nextInt() - 1;
				int x = nextInt();
				arr[k] = x;
				int part = k / ROOT;
				update(part);
			}
			// System.err.println(Arrays.toString(arr));
		}
	}

	D() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new D();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
}