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

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		FastScanner in = new FastScanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskD {
	public void solve(int testNumber, FastScanner in, PrintWriter out) {
		int n = in.nextInt();
		int size = 1;
		while (size < n) {
			size *= 2;
		}
		size *= 2;
		tree = new int[size][T];
		for (int i = 0; i < n; i++) {
			int a = in.nextInt();
			set(0, 0, n - 1, i, a);
		}
		int numQueries = in.nextInt();
		for (int query = 0; query < numQueries; query++) {
			String op = in.next();
			if (op.equals("A")) {
				int l = in.nextInt() - 1;
				int r = in.nextInt() - 2;
				out.println(query(0, 0, n - 1, l, r, 0));
			} else {
				int x = in.nextInt() - 1;
				int y = in.nextInt();
				set(0, 0, n - 1, x, y);
			}
		}
	}


	final int T = 60; // LCM(2, 3, 4, 5, 6)
	int[][] tree;

	private void set(int root, int l, int r, int x, int a) {
		if (l == r) {
			for (int t = 0; t < T; t++) {
				tree[root][t] = t % a == 0 ? 2 : 1;
			}
			return;
		}
		int m = (l + r) / 2;
		if (x <= m) {
			set(2*root + 1, l, m, x, a);
		} else {
			set(2*root + 2, m + 1, r, x, a);
		}
		for (int t = 0; t < T; t++) {
			int u = tree[2*root + 1][t];
			int v = tree[2*root + 2][(t + u) % T];
			tree[root][t] = u + v;
		}
	}

	private int query(int root, int rl, int rr, int l, int r, int x) {
		if (rl > rr || rl > r || l > rr || l > r) {
			return 0;
		}
		if (l == rl && r == rr) {
			return tree[root][x];
		}
		int rm = (rl + rr) / 2;
		int u = query(2*root + 1, rl, rm, l, Math.min(r, rm), x);
		int v = query(2*root + 2, rm + 1, rr, Math.max(l, rm + 1), r, (x + u) % T);
		return u + v;
	}
}

class FastScanner {

	private BufferedReader in;
	private StringTokenizer st;

	public FastScanner(InputStream stream) {
		in = new BufferedReader(new InputStreamReader(stream));
	}

	public String next() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				String rl = in.readLine();
				if (rl == null) {
					return null;
				}
				st = new StringTokenizer(rl);
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return st.nextToken();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

}

