Some country consists of (n + 1) cities, located along a straight highway. Let's number the cities with consecutive integers from 1 to n + 1 in the order they occur along the highway. Thus, the cities are connected by n segments of the highway, the i-th segment connects cities number i and i + 1. Every segment of the highway is associated with a positive integer ai > 1 — the period of traffic jams appearance on it. 

In order to get from city x to city y (x < y), some drivers use the following tactics. 

Initially the driver is in city x and the current time t equals zero. Until the driver arrives in city y, he perfors the following actions:

  * if the current time t is a multiple of ax, then the segment of the highway number x is now having traffic problems and the driver stays in the current city for one unit of time (formally speaking, we assign t = t + 1); 
  * if the current time t is not a multiple of ax, then the segment of the highway number x is now clear and that's why the driver uses one unit of time to move to city x + 1 (formally, we assign t = t + 1 and x = x + 1). 



You are developing a new traffic control system. You want to consecutively process q queries of two types:

  1. determine the final value of time t after the ride from city x to city y (x < y) assuming that we apply the tactics that is described above. Note that for each query t is being reset to 0. 
  2. replace the period of traffic jams appearing on the segment number x by value y (formally, assign ax = y). 



Write a code that will effectively process the queries given above.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of highway segments that connect the n + 1 cities.

The second line contains n integers a1, a2, ..., an (2 ≤ ai ≤ 6) — the periods of traffic jams appearance on segments of the highway.

The next line contains a single integer q (1 ≤ q ≤ 105) — the number of queries to process.

The next q lines contain the descriptions of the queries in the format c, x, y (c — the query type). 

If c is character 'A', then your task is to process a query of the first type. In this case the following constraints are satisfied: 1 ≤ x < y ≤ n + 1.

If c is character 'C', then you need to process a query of the second type. In such case, the following constraints are satisfied: 1 ≤ x ≤ n, 2 ≤ y ≤ 6.

Output

For each query of the first type output a single integer — the final value of time t after driving from city x to city y. Process the queries in the order in which they are given in the input.

Examples

Input

10
2 5 3 2 3 5 3 4 2 4
10
C 10 6
A 2 6
A 1 3
C 3 4
A 3 11
A 4 9
A 5 6
C 7 3
A 8 10
A 2 5


Output

5
3
14
6
2
4
4
import java.io.*;
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * Created by simin on 3/4/15.
 */
public class D
{
	public static int [][]segmentTrees;
	public static final int lcm = 60;
	public static void main(String [] strings) throws IOException
	{
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(reader.readLine());
		StringTokenizer tokenizer = new StringTokenizer(reader.readLine());
		int [] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = Integer.parseInt(tokenizer.nextToken());

		segmentTrees = new int[lcm][4 * n + 7];
		make(0, 0, n, a);

		int q = Integer.parseInt(reader.readLine());
		PrintWriter writer = new PrintWriter(new OutputStreamWriter(System.out));
		for (int i = 0; i < q; i++)
		{
			tokenizer = new StringTokenizer(reader.readLine());
			char c = tokenizer.nextToken().charAt(0);

			if (c == 'A')
			{
				int x = Integer.parseInt(tokenizer.nextToken()) - 1;
				int y = Integer.parseInt(tokenizer.nextToken()) - 1;
				writer.println(get(0, 0, 0, n, x, y, a));
			}
			else
			{
				int x = Integer.parseInt(tokenizer.nextToken()) - 1;
				int y = Integer.parseInt(tokenizer.nextToken());
				if(a[x] == y)
					continue;
				a[x] = y;

				update(0, 0, n, x, a);
			}
		}
		writer.flush();
		writer.close();
	}

	private static void update(int root, int left, int right, int x, int[] a)
	{

		if (x > right || x < left)
			return;
		if (left == right)
		{
			for(int i = 0 ; i < lcm ; i++)
				segmentTrees[i][root] = 0;
			return;
		}
		int mid = (left + right) / 2;
		update(2*root + 1, left, mid, x, a);
		update(2*root + 2, mid+1, right, x, a);
		for(int i = 0 ; i < lcm ; i++)
		{
			int v1 = segmentTrees[i][2 * root + 1];
			int now = (i + v1);
			int h = now % a[mid] == 0 ? 2 : 1;
			int v2 = segmentTrees[(now + h) % lcm][2 * root + 2];
			segmentTrees[i][root] = v1 + h + v2;
		}
	}

	private static int get(int at, int root, int left, int right, int x, int y, int[] a)
	{
		if (left > y || right < x)
			return -1;
		if (left == x && right == y)
		{
			return segmentTrees[at][root];
		}
		int mid = (left + right) / 2;
		int v1 = get(at, 2 * root + 1, left, mid, x, Math.min(mid, y), a);
		int now = (v1 + at);
		int h = now % a[mid] == 0 ? 2 : 1;
		int v2 = get((h + now) % lcm, 2 * root + 2, mid + 1, right, Math.max(x, mid + 1), y, a);
		int res = 0;
		if(v1 != -1)
			res += v1;
		if(v2 != -1)
			res += v2;
		if (v1 != -1 && v2 != -1)
			res += h;
		return res;
	}

	private static void make(int root, int left, int right, int[] a)
	{
		if (left == right)
		{
			for(int i = 0 ; i < lcm ; i++)
				segmentTrees[i][root] = 0;
			return;
		}
		int mid = (left + right) / 2;
		make(2*root + 1, left, mid, a);
		make(2*root + 2, mid+1, right, a);
		for(int i = 0 ; i < lcm ; i++)
		{
			int v1 = segmentTrees[i][2 * root + 1];
			int now = (v1 + i);
			int h = now % a[mid] == 0 ? 2 : 1;
			int v2 = segmentTrees[(now + h) % lcm][2 * root + 2];
			segmentTrees[i][root] = v1 + h + v2;
		}
	}


}