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.math.BigInteger;
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
 * @author shu_mj @ http://shu-mj.com
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskD {
    Scanner in;
    PrintWriter out;

    public void solve(int testNumber, Scanner in, PrintWriter out) {
        this.in = in;
        this.out = out;
        run();
    }

    void run() {
        int n = in.nextInt();
//        int[] is = in.nextIntArray(n);
        Seg seg = new Seg(n);
        int q = in.nextInt();
        while (q-- != 0) {
            char t = in.next().charAt(0);
            int x = in.nextInt() - 1;
            int y = in.nextInt();
            if (t == 'C') {
//                is[x] = y;
                seg.update(x, y);
            } else {
                y--;
//                int crt = 0;
//                for (int i = x; i < y; i++) {
//                    if (crt % is[i] == 0) crt++;
//                    crt++;
//                }
//                out.println(crt);
                out.println(seg.query(x, y));
            }
        }
    }
    class Seg {
        int N;
        int[] as;
        int[] is;

        public Seg(int n) {
            N = Integer.highestOneBit(n) << 1;
            as = new int[N];
            for (int i = 0; i < N; i++) {
                if (i < n) as[i] = in.nextInt();
                else as[i] = 6;
            }
            is = new int[N * 2 * 60];
            for (int i = N * 2 - 1; i > 0; i--) {
                for (int j = 0; j < 60; j++) {
                    if (i < N) {
                        int l = is[i * 2 * 60 + j];
                        is[i * 60 + j] = l + is[(i * 2 + 1) * 60 + (l + j) % 60];
                    } else {
                        if (j % as[i - N] == 0) is[i * 60 + j] = 2;
                        else is[i * 60 + j] = 1;
                    }
                }
            }
        }
        void update(int a, int v) {
            as[a] = v;
            for (int i = a + N; i > 0; i >>= 1) {
                for (int j = 0; j < 60; j++) {
                    if (i < N) {
                        int l = is[i * 2 * 60 + j];
                        is[i * 60 + j] = l + is[(i * 2 + 1) * 60 + (j + l) % 60];
                    } else {
                        if (j % v == 0) is[i * 60 + j] = 2;
                        else is[i * 60 + j] = 1;
                    }
                }
            }
        }
        int[] os = new int[128];
        int query(int s, int t) {
            int l = 0;
            while (0 < s && s + (s & -s) <= t) {
                int i = (N + s) / (s & -s);
                os[l++] = i;
                s += s & -s;
            }
            int r = 128;
            while (s < t) {
                int i = (N + t) / (t & -t) - 1;
                os[--r] = i;
                t -= t & -t;
            }
            int res = 0;
            for (int i = 0; ; i++) {
                if (i == l) i = r;
                if (i >= 128) break;
                res += is[os[i] * 60 + res % 60];
            }
            return res;
        }
    }
}

class Scanner {
    BufferedReader br;
    StringTokenizer st;

    public Scanner(InputStream in) {
        br = new BufferedReader(new InputStreamReader(in));
        eat("");
    }

    private void eat(String s) {
        st = new StringTokenizer(s);
    }

    public String nextLine() {
        try {
            return br.readLine();
        } catch (IOException e) {
            return null;
        }
    }

    public boolean hasNext() {
        while (!st.hasMoreTokens()) {
            String s = nextLine();
            if (s == null)
                return false;
            eat(s);
        }
        return true;
    }

    public String next() {
        hasNext();
        return st.nextToken();
    }

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

}

