#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int n, mn[maxn], q, ans;
struct node {
  int t[65];
  node() { fill_n(t, 65, 0); }
} seg[4 * maxn];
void make(int id, int sl, int sr) {
  if (sr - sl == 1) {
    for (int i = 0; i < 60; i++)
      if (i % mn[sl] == 0)
        seg[id].t[i] = 2;
      else
        seg[id].t[i] = 1;
    return;
  }
  int mid = (sl + sr) / 2;
  make(2 * id, sl, mid);
  make(2 * id + 1, mid, sr);
  for (int i = 0; i < 60; i++)
    seg[id].t[i] =
        seg[2 * id + 1].t[(seg[2 * id].t[i] + i) % 60] + seg[2 * id].t[i];
  return;
}
void upt(int id, int sl, int sr, int i, int val) {
  if (sr - sl == 1) {
    mn[sl] = val;
    for (int j = 0; j < 60; j++)
      if (j % mn[sl] == 0)
        seg[id].t[j] = 2;
      else
        seg[id].t[j] = 1;
    return;
  }
  int mid = (sl + sr) / 2;
  if (mid > i)
    upt(2 * id, sl, mid, i, val);
  else
    upt(2 * id + 1, mid, sr, i, val);
  for (int j = 0; j < 60; j++)
    seg[id].t[j] =
        seg[2 * id + 1].t[(seg[2 * id].t[j] + j) % 60] + seg[2 * id].t[j];
  return;
}
void get(int id, int sl, int sr, int l, int r) {
  int u = max(0, min(r, sr) - max(sl, l));
  if (u == 0) return;
  if (u == sr - sl) {
    ans += seg[id].t[ans % 60];
    return;
  }
  int mid = (sl + sr) / 2;
  get(2 * id, sl, mid, l, r);
  get(2 * id + 1, mid, sr, l, r);
  return;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> mn[i];
  make(1, 0, n);
  cin >> q;
  for (int i = 0; i < q; i++) {
    char type;
    int l, r;
    cin >> type >> l >> r;
    if (type == 'C') {
      l--;
      upt(1, 0, n, l, r);
    } else {
      l--, r--;
      ans = 0;
      get(1, 0, n, l, r);
      cout << ans << endl;
    }
  }
}

### 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;
const int maxn = 1e5 + 10;
int n, mn[maxn], q, ans;
struct node {
  int t[65];
  node() { fill_n(t, 65, 0); }
} seg[4 * maxn];
void make(int id, int sl, int sr) {
  if (sr - sl == 1) {
    for (int i = 0; i < 60; i++)
      if (i % mn[sl] == 0)
        seg[id].t[i] = 2;
      else
        seg[id].t[i] = 1;
    return;
  }
  int mid = (sl + sr) / 2;
  make(2 * id, sl, mid);
  make(2 * id + 1, mid, sr);
  for (int i = 0; i < 60; i++)
    seg[id].t[i] =
        seg[2 * id + 1].t[(seg[2 * id].t[i] + i) % 60] + seg[2 * id].t[i];
  return;
}
void upt(int id, int sl, int sr, int i, int val) {
  if (sr - sl == 1) {
    mn[sl] = val;
    for (int j = 0; j < 60; j++)
      if (j % mn[sl] == 0)
        seg[id].t[j] = 2;
      else
        seg[id].t[j] = 1;
    return;
  }
  int mid = (sl + sr) / 2;
  if (mid > i)
    upt(2 * id, sl, mid, i, val);
  else
    upt(2 * id + 1, mid, sr, i, val);
  for (int j = 0; j < 60; j++)
    seg[id].t[j] =
        seg[2 * id + 1].t[(seg[2 * id].t[j] + j) % 60] + seg[2 * id].t[j];
  return;
}
void get(int id, int sl, int sr, int l, int r) {
  int u = max(0, min(r, sr) - max(sl, l));
  if (u == 0) return;
  if (u == sr - sl) {
    ans += seg[id].t[ans % 60];
    return;
  }
  int mid = (sl + sr) / 2;
  get(2 * id, sl, mid, l, r);
  get(2 * id + 1, mid, sr, l, r);
  return;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> mn[i];
  make(1, 0, n);
  cin >> q;
  for (int i = 0; i < q; i++) {
    char type;
    int l, r;
    cin >> type >> l >> r;
    if (type == 'C') {
      l--;
      upt(1, 0, n, l, r);
    } else {
      l--, r--;
      ans = 0;
      get(1, 0, n, l, r);
      cout << ans << endl;
    }
  }
}

```