#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int maxn = 200010;
int tree[maxn * 2][62], in[maxn];
void merge(int root) {
  int lc = 2 * root;
  int rc = lc + 1;
  for (int i = 0; i < 60; i++) {
    int start_right = (i + tree[lc][i]) % 60;
    int total = tree[lc][i] + tree[rc][start_right];
    tree[root][i] = total;
  }
}
void build(int root, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 60; i++)
      if (i % in[l] == 0)
        tree[root][i] = 2;
      else
        tree[root][i] = 1;
    return;
  }
  int mid = (l + r) / 2;
  build(2 * root, l, mid);
  build(2 * root + 1, mid + 1, r);
  merge(root);
}
void update(int root, int l, int r, int index, int val) {
  if (l == r) {
    in[l] = val;
    for (int i = 0; i < 60; i++)
      if (i % in[l] == 0)
        tree[root][i] = 2;
      else
        tree[root][i] = 1;
    return;
  }
  int mid = (l + r) / 2;
  if (index <= mid)
    update(2 * root, l, mid, index, val);
  else
    update(2 * root + 1, mid + 1, r, index, val);
  merge(root);
}
int query(int root, int x, int y, int l, int r, int tm) {
  if ((x == l) && (y == r)) {
    return tree[root][tm];
  }
  int mid = (l + r) / 2;
  if (y <= mid)
    return query(2 * root, x, y, l, mid, tm);
  else if (x > mid)
    return query(2 * root + 1, x, y, mid + 1, r, tm);
  else {
    int lt = query(2 * root, x, mid, l, mid, tm);
    return lt + query(2 * root + 1, mid + 1, y, mid + 1, r, (lt + tm) % 60);
  }
}
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &in[i]);
  build(1, 1, N);
  int q;
  scanf("%d", &q);
  while (q--) {
    char s[10];
    int x, y;
    scanf("%s %d %d", s, &x, &y);
    if (s[0] == 'C')
      update(1, 1, N, x, y);
    else
      printf("%d\n", query(1, x, y - 1, 1, N, 0));
  }
  return 0;
  return 0;
}

### Prompt

Your task is to create 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 double pi = acos(-1.0);
const double eps = 1e-8;
const int maxn = 200010;
int tree[maxn * 2][62], in[maxn];
void merge(int root) {
  int lc = 2 * root;
  int rc = lc + 1;
  for (int i = 0; i < 60; i++) {
    int start_right = (i + tree[lc][i]) % 60;
    int total = tree[lc][i] + tree[rc][start_right];
    tree[root][i] = total;
  }
}
void build(int root, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 60; i++)
      if (i % in[l] == 0)
        tree[root][i] = 2;
      else
        tree[root][i] = 1;
    return;
  }
  int mid = (l + r) / 2;
  build(2 * root, l, mid);
  build(2 * root + 1, mid + 1, r);
  merge(root);
}
void update(int root, int l, int r, int index, int val) {
  if (l == r) {
    in[l] = val;
    for (int i = 0; i < 60; i++)
      if (i % in[l] == 0)
        tree[root][i] = 2;
      else
        tree[root][i] = 1;
    return;
  }
  int mid = (l + r) / 2;
  if (index <= mid)
    update(2 * root, l, mid, index, val);
  else
    update(2 * root + 1, mid + 1, r, index, val);
  merge(root);
}
int query(int root, int x, int y, int l, int r, int tm) {
  if ((x == l) && (y == r)) {
    return tree[root][tm];
  }
  int mid = (l + r) / 2;
  if (y <= mid)
    return query(2 * root, x, y, l, mid, tm);
  else if (x > mid)
    return query(2 * root + 1, x, y, mid + 1, r, tm);
  else {
    int lt = query(2 * root, x, mid, l, mid, tm);
    return lt + query(2 * root + 1, mid + 1, y, mid + 1, r, (lt + tm) % 60);
  }
}
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &in[i]);
  build(1, 1, N);
  int q;
  scanf("%d", &q);
  while (q--) {
    char s[10];
    int x, y;
    scanf("%s %d %d", s, &x, &y);
    if (s[0] == 'C')
      update(1, 1, N, x, y);
    else
      printf("%d\n", query(1, x, y - 1, 1, N, 0));
  }
  return 0;
  return 0;
}

```