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
{
	public static void main(String [] args) throws IOException
	{
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter writer = new PrintWriter(new OutputStreamWriter(System.out));
		StringTokenizer tokenizer = new StringTokenizer(reader.readLine());

		int n = Integer.parseInt(tokenizer.nextToken());
		int m = Integer.parseInt(tokenizer.nextToken());

		int [] data = new int[n];
		tokenizer = new StringTokenizer(reader.readLine());
		for(int i = 0 ; i < n ; i++)
			data[i] = Integer.parseInt(tokenizer.nextToken());

		MagicTree magicTree = new MagicTree(data);

		while(m-- > 0)
		{
			tokenizer = new StringTokenizer(reader.readLine());
			int type = Integer.parseInt(tokenizer.nextToken());

			switch (type)
			{
				case 1 :
				{
					int left = Integer.parseInt(tokenizer.nextToken()) - 1;
					int right = Integer.parseInt(tokenizer.nextToken()) - 1;

					writer.println(magicTree.getSum(left, right));

					break;
				}
				case 2 :
				{
					int left = Integer.parseInt(tokenizer.nextToken()) - 1;
					int right = Integer.parseInt(tokenizer.nextToken()) - 1;
					int mod = Integer.parseInt(tokenizer.nextToken());

					magicTree.applyMod(left, right, mod);

					break;
				}
				case 3 :
				{
					int idx = Integer.parseInt(tokenizer.nextToken()) - 1;
					int val = Integer.parseInt(tokenizer.nextToken());

					magicTree.setValue(idx, val);

					break;
				}
			}
		}

		writer.flush();
		writer.close();
	}

	static class MagicTree
	{
		private long [] sum;
		private long [] max;
		private int [] data;

		public MagicTree(int [] _data)
		{
			data = _data.clone();
			int n = data.length;

			sum = new long[4*n + 10];
			max = new long[4*n + 10];

			make(0, 0, n-1);
		}

		private void make(int root, int rl, int rr)
		{
			if(rl == rr)
			{
				sum[root] = data[rl];
				max[root] = data[rl];

				return;
			}

			int rm = (rl + rr) >> 1;

			make(2*root+1, rl, rm);
			make(2*root+2, rm+1, rr);

			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 left, int right)
		{
			return internalGetSum(0, 0, data.length-1, left, right);
		}

		private long internalGetSum(int root, int rl, int rr, int left, int right)
		{
			if(left > right)
				return 0;
			if(rl == left && rr == right)
				return sum[root];

			int rm = (rl + rr) >> 1;

			return internalGetSum(2*root+1, rl, rm, left, Math.min(right, rm)) + internalGetSum(2*root+2, rm+1, rr, Math.max(left, rm+1), right);
		}

		private void setValue(int idx, int value)
		{
			internalSetValue(0, 0, data.length-1, idx, value);
		}

		private void internalSetValue(int root, int rl, int rr, int idx, int value)
		{
			if(rl == rr)
			{
				data[idx] = value;
				sum[root] = value;
				max[root] = value;
				return;
			}

			int rm = (rl + rr) >> 1;

			if(idx <= rm)
				internalSetValue(2*root+1, rl, rm, idx, value);
			else
				internalSetValue(2*root+2, rm+1, rr, idx, value);

			sum[root] = sum[2*root+1] + sum[2*root+2];
			max[root] = Math.max(max[2*root+1], max[2*root+2]);
		}

		private void applyMod(int left, int right, int mod)
		{
			internalApplyMode(0, 0, data.length-1, left, right, mod);
		}

		private void internalApplyMode(int root, int rl, int rr, int left, int right, int mod)
		{
			if(left > right)
				return;
			if(max[root] < mod)
				return;
			if(rl == rr)
			{
				data[rl] %= mod;
				sum[root] = data[rl];
				max[root] = data[rl];
				return;
			}

			int rm = (rl + rr) >> 1;
			internalApplyMode(2*root+1, rl, rm, left, Math.min(right, rm), mod);
			internalApplyMode(2*root+2, rm+1, rr, Math.max(left, rm+1), right, mod);

			sum[root] = sum[2*root+1] + sum[2*root+2];
			max[root] = Math.max(max[2*root+1], max[2*root+2]);
		}
	}
}