#include <bits/stdc++.h>
using namespace std;
typedef struct node {
  int base[60];
  node() { memset(base, 0, sizeof base); }
} node;
const int MAX = 1e5 + 5;
node st[4 * MAX];
int n, arr[MAX];
node merge(const node &a, const node &b) {
  node c;
  for (int i = int(0); i < int(60); i++) {
    c.base[i] = a.base[i] + b.base[(i + a.base[i]) % 60];
  }
  return c;
}
void build(int p, int l, int r) {
  if (l == r) {
    for (int i = int(0); i < int(60); i++) {
      st[p].base[i] = !(i % arr[l]) + 1;
    }
    return;
  }
  build(p * 2, l, (l + r) / 2);
  build(p * 2 + 1, (l + r) / 2 + 1, r);
  st[p] = merge(st[p * 2], st[p * 2 + 1]);
}
void update(int p, int l, int r, int idx, int val) {
  if (l == r && l == idx) {
    arr[l] = val;
    for (int i = int(0); i < int(60); i++) {
      st[p].base[i] = !(i % arr[l]) + 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  if (idx <= mid) {
    update(p * 2, l, mid, idx, val);
  } else {
    update(p * 2 + 1, mid + 1, r, idx, val);
  }
  st[p] = merge(st[p * 2], st[p * 2 + 1]);
}
node query(int p, int l, int r, int i, int j) {
  if (l >= i && j >= r) {
    return st[p];
  }
  int mid = (l + r) / 2;
  if (i > r || l > j) {
    return node();
  }
  return merge(query(p * 2, l, mid, i, j), query(p * 2 + 1, mid + 1, r, i, j));
}
int main() {
  scanf("%d%*c", &n);
  for (int i = int(1); i < int(n + 1); i++) {
    scanf("%d%*c", arr + i);
  }
  build(1, 1, n);
  int q;
  scanf("%d%*c", &q);
  for (int i = int(0); i < int(q); i++) {
    char type;
    int x, y;
    scanf("%c %d %d%*c", &type, &x, &y);
    if (type == 'A') {
      y--;
      printf("%d\n", query(1, 1, n, x, y).base[0]);
    } else {
      update(1, 1, n, x, y);
    }
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
typedef struct node {
  int base[60];
  node() { memset(base, 0, sizeof base); }
} node;
const int MAX = 1e5 + 5;
node st[4 * MAX];
int n, arr[MAX];
node merge(const node &a, const node &b) {
  node c;
  for (int i = int(0); i < int(60); i++) {
    c.base[i] = a.base[i] + b.base[(i + a.base[i]) % 60];
  }
  return c;
}
void build(int p, int l, int r) {
  if (l == r) {
    for (int i = int(0); i < int(60); i++) {
      st[p].base[i] = !(i % arr[l]) + 1;
    }
    return;
  }
  build(p * 2, l, (l + r) / 2);
  build(p * 2 + 1, (l + r) / 2 + 1, r);
  st[p] = merge(st[p * 2], st[p * 2 + 1]);
}
void update(int p, int l, int r, int idx, int val) {
  if (l == r && l == idx) {
    arr[l] = val;
    for (int i = int(0); i < int(60); i++) {
      st[p].base[i] = !(i % arr[l]) + 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  if (idx <= mid) {
    update(p * 2, l, mid, idx, val);
  } else {
    update(p * 2 + 1, mid + 1, r, idx, val);
  }
  st[p] = merge(st[p * 2], st[p * 2 + 1]);
}
node query(int p, int l, int r, int i, int j) {
  if (l >= i && j >= r) {
    return st[p];
  }
  int mid = (l + r) / 2;
  if (i > r || l > j) {
    return node();
  }
  return merge(query(p * 2, l, mid, i, j), query(p * 2 + 1, mid + 1, r, i, j));
}
int main() {
  scanf("%d%*c", &n);
  for (int i = int(1); i < int(n + 1); i++) {
    scanf("%d%*c", arr + i);
  }
  build(1, 1, n);
  int q;
  scanf("%d%*c", &q);
  for (int i = int(0); i < int(q); i++) {
    char type;
    int x, y;
    scanf("%c %d %d%*c", &type, &x, &y);
    if (type == 'A') {
      y--;
      printf("%d\n", query(1, 1, n, x, y).base[0]);
    } else {
      update(1, 1, n, x, y);
    }
  }
  return 0;
}

```