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


public class Main {
	static int N = 100005;
	static int n, m, code, x, y, k, mod;
	static long[] a = new long[N];
	static long[] max = new long[N << 2]; //记录某个区间的最大值 若大于要取模的数就不用取模了 否则就一个点一个点进行取模
	static long[] sum = new long[N << 2];
	
	public static void main(String[] args) {
		InputStream inStream = System.in;
		OutputStream outStream = System.out;
		Reader sc = new Reader(inStream);
		PrintWriter out =  new PrintWriter(outStream);
		n = sc.nextInt();
		m = sc.nextInt();
		for (int i = 1; i <= n; i++) {
			a[i] = sc.nextLong();
		}
		build(1, 1, n);
		for (int i = 1; i <= m; i++) {
			code = sc.nextInt();
			if (code == 1) {
				x = sc.nextInt();
				y = sc.nextInt();
				out.println(query(1, 1, n, x, y));
			} else if (code == 2) {
				x = sc.nextInt();
				y = sc.nextInt();
				mod = sc.nextInt();
				update(1, 1, n, x, y, mod);
			} else {
				k = sc.nextInt();
				x = sc.nextInt();
				update(1, 1, n, k, x);
			}
		}
		out.close();
	}
	static void build(int id, int l, int r) {
		if (l == r) {
			sum[id] = max[id] = a[l];
			return;
		}
		int mid = (l + r) >> 1;
		build(id << 1, l, mid);
		build(id << 1 | 1, mid + 1, r);
		pushup(id);
	}
	static void pushup(int id) {
		sum[id] = sum[id << 1] + sum[id << 1 | 1];
		max[id] = max[id << 1] > max[id << 1 | 1] ? max[id << 1] : max[id << 1 | 1];
	}
	static void update(int id, int l, int r, int x, int y, int mod) {
		if (max[id] < mod) {
			return;
		}
		if (l == r) {
			sum[id] = sum[id] % mod;
			max[id] = sum[id];
			return;
		}
		int mid = (l + r) >> 1;
		if (x <= mid) {
			update(id << 1, l, mid, x, y, mod);
		}
		if (y > mid) {
			update(id << 1 | 1, mid + 1, r, x, y, mod);
		}
		pushup(id);
	}
	static void update(int id, int l, int r, int k, int x) {
		if (l == r) {
			sum[id] = max[id] = x;
			return;
		}
		int mid = (l + r) >> 1;
		if (k <= mid) {
			update(id << 1, l, mid, k, x);
		} else {
			update(id << 1 | 1, mid + 1, r, k, x);
		}
		pushup(id);
	}
	static long query(int id, int l, int r, int x, int y) {
		if (x <= l && r <= y) {
			return sum[id];
		}
		int mid = (l + r) >> 1;
		long ans = 0;
		if (x <= mid) {
			ans += query(id << 1, l, mid, x, y);
		}
		if (y > mid) {
			ans += query(id << 1 | 1, mid + 1, r, x, y);
		}
		return ans;
	}
	static class Reader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;
		
		public Reader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}
		
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
	}
}