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
#include <bits/stdc++.h>
int n, q, seg[61][400050], x, y;
int query(int t = 0, int i = 0, int s = 1, int e = -1) {
  if (e == -1) e = n + 1;
  if (x <= s && e <= y) return seg[t][i];
  if (s >= y || e <= x) return 0;
  int tmp = query(t, 2 * i + 1, s, (s + e) / 2);
  return tmp + query((t + tmp) % 60, 2 * i + 2, (s + e) / 2, e);
}
void update(int i = 0, int s = 1, int e = -1) {
  if (e == -1) e = n + 1;
  if (x < s || x >= e) return;
  if (s + 1 == e) {
    for (int t = 0; t < 60; t++) seg[t][i] = t % y == 0 ? 2 : 1;
    return;
  }
  update(2 * i + 1, s, (s + e) / 2);
  update(2 * i + 2, (s + e) / 2, e);
  for (int t = 0; t < 60; t++)
    seg[t][i] =
        seg[t][2 * i + 1] + seg[(t + seg[t][2 * i + 1]) % 60][2 * i + 2];
}
int main() {
  scanf("%d\n", &n);
  for (int i = 0; i < n; i++) {
    x = i + 1;
    scanf("%d", &y);
    update();
  }
  scanf("%d\n", &q);
  for (int i = 0; i < q; i++) {
    char c;
    scanf("%c %d %d\n", &c, &x, &y);
    if (c == 'A')
      printf("%d\n", query());
    else
      update();
  }
  return 0;
}
