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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Objects;
import java.util.Optional;
import java.util.StringTokenizer;
import java.util.function.BiConsumer;
import java.util.function.BiFunction;
import java.util.function.Function;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.util.stream.LongStream;
import java.util.stream.Stream;

public class Solution {
    static int n, q;
    static final int HEAP_SIZE = 1 << 17;
    static int[][] heaps = new int[60][HEAP_SIZE * 2];
    static boolean[] valid = new boolean[HEAP_SIZE * 2];
    static int[] periods = new int[HEAP_SIZE];

    static void invalidate(int road) {
        for (int i = road + HEAP_SIZE; i > 0; i /= 2) {
            valid[i] = false;
        }
    }

    static void fix(int node) {
        valid[node] = true;
        if (node >= HEAP_SIZE) {
            for (int time = 0; time < 60; time++) {
                heaps[time][node] = time % periods[node - HEAP_SIZE] == 0 ? 2 : 1;
            }
            return;
        }
        for (int i = 0; i < 2; i++) {
            if (!valid[2 * node + i]) {
                fix(2 * node + i);
            }
        }
        for (int time = 0; time < 60; time++) {
            int leftTime = heaps[time][2 * node];
            int rightTime = heaps[(time + leftTime) % 60][2 * node + 1];
            heaps[time][node] = leftTime + rightTime;
        }
    }

    static int travel(int from, int to, int initialTime, int node, int left, int right) {
        if (from == left && to == right) {
            if (!valid[node]) {
                fix(node);
            }
            return heaps[initialTime % 60][node];
        }
        int middle = (left + right) / 2;
        if (from >= middle) {
            return travel(from, to, initialTime, 2 * node + 1, middle, right);
        }
        if (to <= middle) {
            return travel(from, to, initialTime, 2 * node, left, middle);
        }
        int leftTime = travel(from, middle, initialTime, 2 * node, left, middle);
        int rightTime = travel(middle, to, (initialTime + leftTime) % 60, 2 * node + 1, middle, right);
        return leftTime + rightTime;
    }

    public static void main(String[] args) throws IOException {
        try (BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
             PrintWriter writer = new PrintWriter(System.out)) {
            n = nextInt(reader);
            for (int i = 0; i < n; i++) {
                periods[i] = nextInt(reader);
            }
            q = nextInt(reader);
            for (int k = 0; k < q; k++) {
                if (next(reader).charAt(0) == 'A') {
                    writer.println(travel(nextInt(reader) - 1, nextInt(reader) - 1, 0, 1, 0, HEAP_SIZE));
                } else {
                    int city = nextInt(reader) - 1;
                    invalidate(city);
                    periods[city] = nextInt(reader);
                }
            }
        }
    }

    private static StringTokenizer stringTokenizer;

    private static String next(BufferedReader reader) throws IOException {
        if (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
            stringTokenizer = new StringTokenizer(reader.readLine());
        }
        return stringTokenizer.nextToken();
    }

    private static int nextInt(BufferedReader reader) throws IOException {
        return Integer.parseInt(next(reader));
    }

    private static long nextLong(BufferedReader reader) throws IOException {
        return Long.parseLong(next(reader));
    }
}
