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>
const int N = 100005;
using namespace std;
int n, a[N], t[N * 4][60], q, x, y, s;
void build(int k, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 60; ++i) t[k][i] = (i % a[l] == 0 ? 1 : 0) + 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(k * 2, l, mid);
  build(k * 2 + 1, mid + 1, r);
  for (int i = 0; i < 60; ++i)
    t[k][i] = t[k * 2][i] + t[k * 2 + 1][(i + t[k * 2][i]) % 60];
}
int query(int k, int l, int r) {
  int res = 0;
  if (x <= l && r <= y) {
    res = t[k][s];
    s = (s + t[k][s]) % 60;
    return res;
  }
  int mid = (l + r) >> 1;
  if (x <= mid) res += query(k * 2, l, mid);
  if (y > mid) res += query(k * 2 + 1, mid + 1, r);
  return res;
}
void modify(int k, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 60; ++i) t[k][i] = (i % y == 0 ? 1 : 0) + 1;
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    modify(k * 2, l, mid);
  else
    modify(k * 2 + 1, mid + 1, r);
  for (int i = 0; i < 60; ++i)
    t[k][i] = t[k * 2][i] + t[k * 2 + 1][(i + t[k * 2][i]) % 60];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  build(1, 1, n);
  scanf("%d", &q);
  char ss[5];
  while (q-- > 0) {
    scanf("%s", ss);
    char c = ss[0];
    scanf("%d%d", &x, &y);
    if (c == 'A') --y, s = 0, printf("%d\n", query(1, 1, n));
    if (c == 'C') modify(1, 1, n);
  }
  return 0;
}
