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

/**
 * 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;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskD {
    private int n, q;
    private int[] a;
    private int[][] it;

    private void Build(int i, int l, int r) {
        if (l == r) {
            for (int j = 0; j < 60; ++j)
                it[j][i] = (j % a[l] == 0 ? 2 : 1);
            return;
        }
        int mid = (l + r) >> 1;
        Build(i + i, l, mid);
        Build(i + i + 1, mid + 1, r);
        for (int time = 0; time < 60; ++time)
            it[time][i] = it[time][i + i] + it[(time + it[time][i + i]) % 60][i + i + 1];
    }

    private int Get(int time, int i, int l, int r, int u, int v) {
        if (l > v || r < u) return 0;
        if (l >= u && r <= v) return it[time][i];
        int mid = (l + r) >> 1;
        int left = Get(time, i + i, l, mid, u, v);
        int right = Get((time + left) % 60, i + i + 1, mid + 1, r, u, v);
        return left + right;
    }

    private void Update(int i, int l, int r, int u, int v) {
        if (l > u || r < u) return;
        if (l == r) {
            a[l] = v;
            for (int j = 0; j < 60; ++j)
                it[j][i] = (j % a[l] == 0 ? 2 : 1);
            return;
        }
        int mid = (l + r) >> 1;
        Update(i + i, l, mid, u, v);
        Update(i + i + 1, mid + 1, r, u, v);
        for (int j = 0; j < 60; ++j)
            it[j][i] = it[j][i + i] + it[(j + it[j][i + i]) % 60][i + i + 1];
    }

    public void solve(int testNumber, InputReader in, PrintWriter out) {
        n = in.nextInt();
        a = new int[n + 1];
        it = new int[60][4 * n + 5];

        for (int i = 1; i <= n; ++i) a[i] = in.nextInt();

        Build(1, 1, n);

        q = in.nextInt();
        while (q-- > 0) {
            String ch = in.next();
            if (ch.equals("A")) {
                int x = in.nextInt(), y = in.nextInt();
                out.println(Get(0, 1, 1, n, x, y - 1));
            } else {
                int x = in.nextInt(), y = in.nextInt();
                Update(1, 1, n, x, y);
            }
        }
    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
        tokenizer = null;
    }

    public String next() {
        try {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(reader.readLine());
            }
            return tokenizer.nextToken();
        } catch (Exception e) {
            throw new UnknownError(e.getMessage());
        }
    }

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

}

