#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
struct node {
  int d[60];
  node operator+(const node& t) const {
    node tmp;
    for (int i = 0; i < 60; i++)
      tmp.d[i] = this->d[i] + t.d[(i + (this->d[i])) % 60];
    return tmp;
  }
} a[maxn * 4];
void up(int k, int l, int r, int p, int val) {
  if (l == r) {
    for (int i = 0; i < 60; i++) {
      if (i % val == 0)
        a[k].d[i] = 2;
      else
        a[k].d[i] = 1;
    }
  } else {
    int m = (l + r) / 2;
    if (p <= m)
      up(k << 1, l, m, p, val);
    else if (p > m)
      up(k << 1 | 1, m + 1, r, p, val);
    a[k] = a[k << 1] + a[k << 1 | 1];
  }
}
node qu(int k, int l, int r, int L, int R) {
  if (l >= L && r <= R) return a[k];
  int m = (l + r) / 2;
  if (R <= m)
    return qu(k << 1, l, m, L, R);
  else if (L > m)
    return qu(k << 1 | 1, m + 1, r, L, R);
  return qu(k << 1, l, m, L, R) + qu(k << 1 | 1, m + 1, r, L, R);
}
int main() {
  int n, q;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    up(1, 1, n, i, x);
  }
  cin >> q;
  while (q--) {
    char aa;
    int bb, cc;
    cin >> aa >> bb >> cc;
    if (aa == 'A') {
      printf("%d\n", qu(1, 1, n, bb, cc - 1).d[0]);
    } else {
      up(1, 1, n, bb, cc);
    }
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
struct node {
  int d[60];
  node operator+(const node& t) const {
    node tmp;
    for (int i = 0; i < 60; i++)
      tmp.d[i] = this->d[i] + t.d[(i + (this->d[i])) % 60];
    return tmp;
  }
} a[maxn * 4];
void up(int k, int l, int r, int p, int val) {
  if (l == r) {
    for (int i = 0; i < 60; i++) {
      if (i % val == 0)
        a[k].d[i] = 2;
      else
        a[k].d[i] = 1;
    }
  } else {
    int m = (l + r) / 2;
    if (p <= m)
      up(k << 1, l, m, p, val);
    else if (p > m)
      up(k << 1 | 1, m + 1, r, p, val);
    a[k] = a[k << 1] + a[k << 1 | 1];
  }
}
node qu(int k, int l, int r, int L, int R) {
  if (l >= L && r <= R) return a[k];
  int m = (l + r) / 2;
  if (R <= m)
    return qu(k << 1, l, m, L, R);
  else if (L > m)
    return qu(k << 1 | 1, m + 1, r, L, R);
  return qu(k << 1, l, m, L, R) + qu(k << 1 | 1, m + 1, r, L, R);
}
int main() {
  int n, q;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    up(1, 1, n, i, x);
  }
  cin >> q;
  while (q--) {
    char aa;
    int bb, cc;
    cin >> aa >> bb >> cc;
    if (aa == 'A') {
      printf("%d\n", qu(1, 1, n, bb, cc - 1).d[0]);
    } else {
      up(1, 1, n, bb, cc);
    }
  }
  return 0;
}

```