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.*;
import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = s.nextInt();
		}
		Node root = new Node(0, n - 1, a);
		int q = s.nextInt();
		s.nextLine();
		while (q-- > 0) {
			StringTokenizer st = new StringTokenizer(s.nextLine());
			String c = st.nextToken();
			int x = Integer.parseInt(st.nextToken()), y = Integer.parseInt(st.nextToken());
			if (c.equals("A")) {
				System.out.println(root.getCost(0, x - 1, y - 2));
			} else {
				root.update(x - 1, y);
			}
		}
	}

	public static class Node {
		public Node left, right;
		public int[] cost;
		public int l, r;

		public Node(int l, int r, int[] a) {
			this.l = l;
			this.r = r;
			cost = new int[M];
			if (l == r) {
				int v = a[l];
				for (int i = 0; i < M; i++) {
					if (i % v == 0) {
						cost[i] = 2;
					} else {
						cost[i] = 1;
					}
				}
			} else {
				int mid = (l + r) >> 1;
				left = new Node(l, mid, a);
				right = new Node(mid + 1, r, a);
				for (int i = 0; i < M; i++) {
					cost[i] = left.cost[i] + right.cost[(i + left.cost[i]) % M];
				}
			}
		}

		public void update(int x, int y) {
			if (l <= x && x <= r) {
				if (l == r) {
					for (int i = 0; i < M; i++) {
						if (i % y == 0) {
							cost[i] = 2;
						} else {
							cost[i] = 1;
						}
					}
				} else {
					left.update(x, y);
					right.update(x, y);
					for (int i = 0; i < M; i++) {
						cost[i] = left.cost[i] + right.cost[(i + left.cost[i]) % M];
					}
				}
			}
		}

		public int getCost(int t, int x, int y) {
			if (y < l || x > r) {
				return 0;
			}
			if (x <= l && r <= y) {
				return cost[t % M];
			}
			int cl = left.getCost(t, x, y);
			return cl + right.getCost(t + cl, x, y);
		}
	}
	public static final int M = 60;
}
