#include <bits/stdc++.h>
using namespace std;
int n, a[100005], cnt[100005][60];
void upd(int x) {
  for (int i = 59; i >= 0; i--) cnt[x][i] = 1 + (i % a[x] == 0);
  x--;
  while ((x + 1) % 40 != 0) {
    for (int i = 59; i >= 0; i--)
      cnt[x][i] = 1 + cnt[x + (i % a[x] != 0)][(i + 1) % 60];
    x--;
  }
}
int main() {
  int q;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (i == n - 1 || (i + 1) % 40 == 0) upd(i);
  }
  scanf("%d", &q);
  while (q--) {
    char t;
    int l, r;
    scanf(" %c%d%d", &t, &l, &r);
    l--;
    if (t == 'C') {
      a[l] = r;
      upd(min(l - l % 40 + 40, n) - 1);
    } else {
      r--;
      int ct = 0;
      while (1) {
        int nex = l - l % 40 + 40;
        if (nex >= r) break;
        ct += cnt[l][ct % 60];
        l = nex;
      }
      while (l != r) {
        if (ct % a[l]) l++;
        ct++;
      }
      printf("%d\n", ct);
    }
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, a[100005], cnt[100005][60];
void upd(int x) {
  for (int i = 59; i >= 0; i--) cnt[x][i] = 1 + (i % a[x] == 0);
  x--;
  while ((x + 1) % 40 != 0) {
    for (int i = 59; i >= 0; i--)
      cnt[x][i] = 1 + cnt[x + (i % a[x] != 0)][(i + 1) % 60];
    x--;
  }
}
int main() {
  int q;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (i == n - 1 || (i + 1) % 40 == 0) upd(i);
  }
  scanf("%d", &q);
  while (q--) {
    char t;
    int l, r;
    scanf(" %c%d%d", &t, &l, &r);
    l--;
    if (t == 'C') {
      a[l] = r;
      upd(min(l - l % 40 + 40, n) - 1);
    } else {
      r--;
      int ct = 0;
      while (1) {
        int nex = l - l % 40 + 40;
        if (nex >= r) break;
        ct += cnt[l][ct % 60];
        l = nex;
      }
      while (l != r) {
        if (ct % a[l]) l++;
        ct++;
      }
      printf("%d\n", ct);
    }
  }
}

```