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.math.BigInteger;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author AlexFetisov
 */
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 {

    final static int LCM = 60;

    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int[] a = Utils.readIntArray(in, n);
        IntervalTree tree = new IntervalTree(n);
        for (int i = 0; i < n; ++i) {
            tree.set(i, a[i]);
        }

        int countQueries = in.nextInt();
        while (countQueries-- > 0) {
            char c = in.nextString().charAt(0);
            int x = in.nextInt()-1;
            int y = in.nextInt();
            if (c == 'A') {
                --y;
                --y;
                out.println(tree.get(x, y));
            } else {
                tree.set(x, y);
            }
        }
    }

    class IntervalTree {
        int maxTree;
        int n;

        int[][] delta;

        public IntervalTree(int size) {
            n = size;
            maxTree = 1;
            while (maxTree <= n) maxTree <<= 1;
            maxTree <<= 1;
            delta = new int[maxTree][LCM];
        }

        int internalGet(int v, int rangeLeft, int rangeRight, int left, int right, int enter) {
            if (rangeLeft == left && rangeRight == right) {
                return delta[v][enter];
            }
            int middle = (rangeLeft + rangeRight) >> 1;
            int res = 0;
            if (left <= middle) {
                res = internalGet(v << 1, rangeLeft, middle, left, Math.min(right, middle), enter);
                enter = (enter + res) % LCM;
            }
            if (right > middle) {
                res += internalGet(v * 2 + 1, middle + 1, rangeRight, Math.max(left, middle+1), right, enter);
            }
            return res;
        }

        int get(int left, int right) {
            return internalGet(1, 0, n-1, left, right, 0);
        }

        void set(int pos, int x) {
            int v = 1;
            int rangeLeft = 0, rangeRight = n-1;
            while (rangeLeft < rangeRight) {
                int middle = (rangeLeft + rangeRight) / 2;
                v <<= 1;
                if (pos > middle) {
                    ++v;
                    rangeLeft = middle + 1;
                } else {
                    rangeRight = middle;
                }
            }
            for (int i = 0; i < LCM; ++i) {
                if (i % x == 0) {
                    delta[v][i] = 2;
                } else {
                    delta[v][i] = 1;
                }
            }
            v >>= 1;
            while (v != 0) {
                for (int i = 0; i < LCM; ++i) {
                    int d = delta[v<<1][i];
                    int ni = (i + d) % LCM;
                    d += delta[v*2+1][ni];
                    delta[v][i] = d;
                }
                v >>= 1;
            }
        }
    }
}

class InputReader {
    private BufferedReader reader;
    private StringTokenizer stt;

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

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

    public String nextString() {
        while (stt == null || !stt.hasMoreTokens()) {
            stt = new StringTokenizer(nextLine());
        }
        return stt.nextToken();
    }

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

}

class Utils {
    public static int[] readIntArray(InputReader in, int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; ++i) {
            a[i] = in.nextInt();
        }
        return a;
    }

}

