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.util.Scanner;

public class main4 {
	static class PerfectTree {
		private int[] data;
		private long[] sum;
		private int[] max;
		private int n;
		
		public PerfectTree(int[] data) {
			this.data = data;
			n = data.length;
			sum = new long[5 * n];
			max = new int[5 * n];
			
			buildRec( 0, 0, n - 1);
		}
		
		private void buildRec( int vertex, int l, int r) {
			if (l > r)
				return;
			
			if (l == r) {
				max[vertex] = data[l];
				sum[vertex] = data[l];
				return;
			}
			
			int mid = (l + r) >> 1;
			int vl = 2*vertex + 1;
			int vr  =2*vertex + 2;
			buildRec( vl, l, mid);
			buildRec( vr, mid + 1, r);
			
			sum[vertex] = sum[vl] + sum[vr];
			max[vertex] = Math.max(max[vl], max[vr]);
			
		}
		
		private long getInternalSum(int vertex, int l, int r, int origL, int origR) {
			if (l > r) 
				return 0;
			
			if (l > origR || r < origL) {
				return 0;
			}
			
			long ans = 0;
			
			if (l == r)
				return sum[vertex];
			
			if(l>=origL &&r<=origR)
				return sum[vertex];
			
			int mid = (l + r) >> 1;
			
			long ans1 = getInternalSum(2*vertex + 1, l, mid, origL, origR);
			long ans2 = getInternalSum(2*vertex + 2, mid + 1, r, origL, origR);
			
			ans = ans1 + ans2;
			
			return ans;
		}
		
		private void applyInternalMod(int vertex, int l, int r, int origL, int origR, int x) {
			if (l > r) 
				return;
			
			if (l > origR || r < origL) {
				return;
			}
			
			if(max[vertex] < x)
			{
				return;
			}
			
			if (l == r) {
				data[l] %= x;
				max[vertex] = data[l];
				sum[vertex] = data[l];
				
				return ;
			}
			
			int mid = (l + r) >> 1;
			int vl = 2*vertex + 1;
			int vr  =2*vertex + 2;
			applyInternalMod(vl, l, mid, origL, origR, x);
			applyInternalMod(vr, mid + 1, r, origL, origR, x);
 			
			max[vertex] = Math.max(max[vl], max[vr]);
			sum[vertex] = sum[vl] + sum[vr];
		}
		
		private void setElementInternal(int vertex, int l, int r, int k, int x) {
			if (l > r) 
				return;
			
			if (k < l || r < k)
				return;
			
			if ( l == r && r == k) {
				data[l] = x;
				max[vertex] = data[l];
				sum[vertex] = data[l];
				
				return ;
			}
			
			int mid = (l + r) >> 1;
			int vl = 2*vertex + 1;
			int vr  =2*vertex + 2;
			if (k <= mid)
				setElementInternal(vl, l, mid, k, x);
			else
				setElementInternal(vr, mid + 1, r, k, x);
			
			max[vertex] = Math.max(max[vl], max[vr]);
			sum[vertex] = sum[vl] + sum[vr];
		}
		
		public long getSum(int l, int r) {
			return getInternalSum(0, 0, data.length - 1, l , r);
		}
		
		public void applyMod(int l, int r, int x) {
			applyInternalMod(0, 0, n - 1, l, r, x);
		}
		
		public void setElement(int k, int x) {
			setElementInternal(0, 0, n - 1, k, x);
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		
		int[] data =new int[n];
		sc.nextLine();
		String line = sc.nextLine();
		String[] vals = line.split(" ");
		for (int i = 0; i < data.length; i++) {
			data[i] = Integer.parseInt(vals[i]);
		}
		
		PerfectTree tree = new PerfectTree(data);
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < m; i++) {
			int op = sc.nextInt();
			
			if (op == 1) {
				int l = sc.nextInt() - 1;
				int r = sc.nextInt() - 1;
				
				long sum = tree.getSum(l, r);
				sb.append(sum);
				sb.append("\n");
			} else if (op == 2) {
				int l = sc.nextInt() - 1;
				int r = sc.nextInt() - 1;
				int x = sc.nextInt();
				
				tree.applyMod(l, r, x);
			} else {
				int k = sc.nextInt() - 1;
				int x = sc.nextInt();
				
				tree.setElement(k, x);
			}
		}
		
		System.out.println(sb.toString());
	}
}