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
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF498D extends PrintWriter {
	CF498D() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF498D o = new CF498D(); o.main(); o.flush();
	}

	int[][] tr; int[] qq; int n;
	void pul(int i) {
		int l = i << 1, r = l | 1;
		for (int h = 0; h < 60; h++) {
			int t = tr[l][h];
			tr[i][h] = t + tr[r][(h + t) % 60];
		}
	}
	void put(int i, int a) {
		for (int h = 0; h < 60; h++)
			tr[i][h] = h % a == 0 ? 2 : 1;
	}
	void build(int[] aa) {
		for (int i = 0; i < n; i++)
			put(n + i, aa[i]);
		for (int i = n - 1; i > 0; i--)
			pul(i);
	}
	void update(int i, int a) {
		put(i += n, a);
		while (i > 1)
			pul(i >>= 1);
	}
	int query(int l, int r) {
		int cnt = 0, t = 0;
		for (l += n, r += n; l <= r; l >>= 1, r >>= 1) {
			if ((l & 1) == 1)
				t += tr[l++][t % 60];
			if ((r & 1) == 0)
				qq[cnt++] = r--;
		}
		while (cnt-- > 0)
			t += tr[qq[cnt]][t % 60];
		return t;
	}
	void main() {
		n = sc.nextInt();
		int[] aa = new int[n];
		for (int i = 0; i < n; i++)
			aa[i] = sc.nextInt();
		tr = new int[n * 2][60];
		qq = new int[n];
		build(aa);
		int q = sc.nextInt();
		while (q-- > 0) {
			String s = sc.next();
			if (s.charAt(0) == 'C') {
				int i = sc.nextInt() - 1;
				int a = sc.nextInt();
				update(i, a);
			} else {
				int l = sc.nextInt() - 1;
				int r = sc.nextInt() - 1;
				println(query(l, r - 1));
			}
		}
	}
}
