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.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class CF284D {
    static int[] a;
    static int[] tree;

    static void build(int i, int l, int r) {
        if (l == r) {
            for (int k = 0; k < 60; k++) {
                if (k % a[l] == 0)
                    tree[i * 60 + k] = 2;
                else
                    tree[i * 60 + k] = 1;
            }
            return;
        }
        build(i * 2 + 1, l, (l + r) / 2);
        build(i * 2 + 2, (l + r) / 2 + 1, r);
        for (int k = 0; k < 60; k++) {
            int left = tree[(i * 2 + 1) * 60 + k];
            int right = tree[(i * 2 + 2) * 60 + (left + k) % 60];
            tree[i * 60 + k] = left + right;
        }

    }

    static int t = 0;

    static void query(int i, int a, int b, int l, int r) {
        if (b < l || a > r)
            return;
        if (a >= l && b <= r) {
            t += tree[i * 60 + t % 60];
            return;
        }
        query(i * 2 + 1, a, (a + b) / 2, l, r);
        query(i * 2 + 2, (a + b) / 2 + 1, b, l, r);
    }

    static void update(int i, int l, int r, int target) {
        if (l == r) {
            for (int k = 0; k < 60; k++) {
                if (k % a[target] == 0)
                    tree[i * 60 + k] = 2;
                else
                    tree[i * 60 + k] = 1;
            }
            return;
        }
        if (target <= (l + r >> 1))
            update(i * 2 + 1, l, l + r >> 1, target);
        else
            update(i * 2 + 2, (l + r >> 1) + 1, r, target);
        for (int k = 0; k < 60; k++) {
            int left = tree[(i * 2 + 1) * 60 + k];
            int right = tree[(i * 2 + 2) * 60 + (left + k) % 60];
            tree[i * 60 + k] = left + right;
        }
    }

    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer strtok;
        strtok = new StringTokenizer(in.readLine());
        int n = Integer.parseInt(strtok.nextToken());
        a = new int[n];
        tree = new int[n * 4 * 60];
        strtok = new StringTokenizer(in.readLine());
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(strtok.nextToken());
        }
        build(0, 0, n - 1);
        strtok = new StringTokenizer(in.readLine());
        int m = Integer.parseInt(strtok.nextToken());
        for (int q = 0; q < m; q++) {
            strtok = new StringTokenizer(in.readLine());
            String type = strtok.nextToken();
            if (type.equals("A")) {
                int x = Integer.parseInt(strtok.nextToken()) - 1;
                int y = Integer.parseInt(strtok.nextToken()) - 1;
                if (x == y) {
                    System.out.println(0);
                } else {
                    t = 0;
                    query(0, 0, n - 1, x, y - 1);
                    System.out.println(t);
                }
            } else {
                int x = Integer.parseInt(strtok.nextToken()) - 1;
                int y = Integer.parseInt(strtok.nextToken());
                a[x] = y;
                update(0, 0, n - 1, x);
            }
        }
    }
}
/*
 * 100 4 6 4 4 5 5 2 6 4 5 3 4 2 6 2 6 2 3 3 6 5 4 4 6 6 2 4 3 3 6 3 4 2 2 2 4 4
 * 4 2 5 5 2 3 4 5 6 6 6 6 6 3 5 3 3 5 6 6 4 3 5 4 6 6 5 6 2 2 4 2 6 3 3 4 5 2 2
 * 5 4 2 6 6 4 6 6 3 4 3 3 2 3 5 3 3 6 5 3 6 5 6 4 1 A 7 37
 */
