#include <bits/stdc++.h>
using namespace std;
const int nmax = 1e5 + 42;
int tree[nmax * 4][60];
int n;
int arr[nmax];
int t;
void make_node(int node, int l) {
  for (int i = 0; i < 60; i++)
    if (i % arr[l] == 0)
      tree[node][i] = 2;
    else
      tree[node][i] = 1;
}
void combine(int node) {
  for (int i = 0; i < 60; i++)
    tree[node][i] =
        tree[node * 2][i] + tree[node * 2 + 1][(i + tree[node * 2][i]) % 60];
}
void build(int node, int l, int r) {
  if (l == r) {
    make_node(node, l);
    return;
  }
  int av = (l + r) / 2;
  build(node * 2, l, av);
  build(node * 2 + 1, av + 1, r);
  combine(node);
}
void update(int node, int l, int r, int pos) {
  if (l == r) {
    make_node(node, l);
    return;
  }
  int av = (l + r) / 2;
  if (pos <= av)
    update(node * 2, l, av, pos);
  else
    update(node * 2 + 1, av + 1, r, pos);
  combine(node);
}
void go(int node, int l, int r, int lq, int rq) {
  if (l == lq && r == rq) {
    t = t + tree[node][t % 60];
    return;
  }
  int av = (l + r) / 2;
  if (lq <= av) go(node * 2, l, av, lq, min(rq, av));
  if (av < rq) go(node * 2 + 1, av + 1, r, max(av + 1, lq), rq);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  build(1, 1, n);
  int q;
  cin >> q;
  char type;
  int x, y;
  for (int i = 1; i <= q; i++) {
    cin >> type >> x >> y;
    if (type == 'C') {
      arr[x] = y;
      update(1, 1, n, x);
    } else {
      t = 0;
      go(1, 1, n, x, y - 1);
      cout << t << "\n";
    }
  }
  cout << endl;
  return 0;
}

### 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 nmax = 1e5 + 42;
int tree[nmax * 4][60];
int n;
int arr[nmax];
int t;
void make_node(int node, int l) {
  for (int i = 0; i < 60; i++)
    if (i % arr[l] == 0)
      tree[node][i] = 2;
    else
      tree[node][i] = 1;
}
void combine(int node) {
  for (int i = 0; i < 60; i++)
    tree[node][i] =
        tree[node * 2][i] + tree[node * 2 + 1][(i + tree[node * 2][i]) % 60];
}
void build(int node, int l, int r) {
  if (l == r) {
    make_node(node, l);
    return;
  }
  int av = (l + r) / 2;
  build(node * 2, l, av);
  build(node * 2 + 1, av + 1, r);
  combine(node);
}
void update(int node, int l, int r, int pos) {
  if (l == r) {
    make_node(node, l);
    return;
  }
  int av = (l + r) / 2;
  if (pos <= av)
    update(node * 2, l, av, pos);
  else
    update(node * 2 + 1, av + 1, r, pos);
  combine(node);
}
void go(int node, int l, int r, int lq, int rq) {
  if (l == lq && r == rq) {
    t = t + tree[node][t % 60];
    return;
  }
  int av = (l + r) / 2;
  if (lq <= av) go(node * 2, l, av, lq, min(rq, av));
  if (av < rq) go(node * 2 + 1, av + 1, r, max(av + 1, lq), rq);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  build(1, 1, n);
  int q;
  cin >> q;
  char type;
  int x, y;
  for (int i = 1; i <= q; i++) {
    cin >> type >> x >> y;
    if (type == 'C') {
      arr[x] = y;
      update(1, 1, n, x);
    } else {
      t = 0;
      go(1, 1, n, x, y - 1);
      cout << t << "\n";
    }
  }
  cout << endl;
  return 0;
}

```