#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10, G = 60;
int seg[4 * MAXN][G];
int a[MAXN];
inline void pull(int x) {
  int l = 2 * x, r = l + 1;
  for (int i = 0; i < G; i++) {
    int next = (seg[l][i] + i) % G;
    seg[x][i] = seg[r][next] + seg[l][i];
  }
}
void upd(int x, int xl, int xr, int ind, int q) {
  if (ind >= xr || ind < xl) return;
  if (xr - xl == 1) {
    a[xl] = q;
    for (int i = 0; i < G; i++) {
      seg[x][i] = 1;
      if (i % a[xl] == 0) seg[x][i]++;
    }
    return;
  }
  int mid = (xr + xl) / 2, l = 2 * x, r = l + 1;
  upd(l, xl, mid, ind, q);
  upd(r, mid, xr, ind, q);
  pull(x);
}
int get(int x, int xl, int xr, int ql, int qr, int q) {
  if (xl >= qr || xr <= ql) return 0;
  if (xl >= ql && xr <= qr) {
    return seg[x][q];
  }
  int mid = (xl + xr) / 2, l = 2 * x, r = l + 1;
  int u = get(l, xl, mid, ql, qr, q), v = get(r, mid, xr, ql, qr, (u + q) % G);
  return u + v;
}
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    upd(1, 0, n, i, a[i]);
  }
  int m;
  cin >> m;
  while (m--) {
    char com;
    int l, r;
    cin >> com >> l >> r;
    if (com == 'A') {
      l--;
      r--;
      cout << get(1, 0, n, l, r, 0) << "\n";
    } else {
      l--;
      upd(1, 0, n, l, r);
    }
  }
}

### Prompt

Please formulate a Cpp solution to the following problem:

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;
const int MAXN = 1e5 + 10, G = 60;
int seg[4 * MAXN][G];
int a[MAXN];
inline void pull(int x) {
  int l = 2 * x, r = l + 1;
  for (int i = 0; i < G; i++) {
    int next = (seg[l][i] + i) % G;
    seg[x][i] = seg[r][next] + seg[l][i];
  }
}
void upd(int x, int xl, int xr, int ind, int q) {
  if (ind >= xr || ind < xl) return;
  if (xr - xl == 1) {
    a[xl] = q;
    for (int i = 0; i < G; i++) {
      seg[x][i] = 1;
      if (i % a[xl] == 0) seg[x][i]++;
    }
    return;
  }
  int mid = (xr + xl) / 2, l = 2 * x, r = l + 1;
  upd(l, xl, mid, ind, q);
  upd(r, mid, xr, ind, q);
  pull(x);
}
int get(int x, int xl, int xr, int ql, int qr, int q) {
  if (xl >= qr || xr <= ql) return 0;
  if (xl >= ql && xr <= qr) {
    return seg[x][q];
  }
  int mid = (xl + xr) / 2, l = 2 * x, r = l + 1;
  int u = get(l, xl, mid, ql, qr, q), v = get(r, mid, xr, ql, qr, (u + q) % G);
  return u + v;
}
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    upd(1, 0, n, i, a[i]);
  }
  int m;
  cin >> m;
  while (m--) {
    char com;
    int l, r;
    cin >> com >> l >> r;
    if (com == 'A') {
      l--;
      r--;
      cout << get(1, 0, n, l, r, 0) << "\n";
    } else {
      l--;
      upd(1, 0, n, l, r);
    }
  }
}

```