#include <bits/stdc++.h>
using namespace std;
const int maxN = 101000;
int t[4 * maxN][60];
int a[maxN];
int get(int v, int cl, int cr, int l, int r, int mod) {
  if (cl == l && cr == r) {
    return t[v][mod];
  }
  int mid = (cl + cr) / 2;
  if (r <= mid) {
    return get(2 * v, cl, mid, l, r, mod);
  } else if (l > mid) {
    return get(2 * v + 1, mid + 1, cr, l, r, mod);
  } else {
    int v1 = get(2 * v, cl, mid, l, mid, mod);
    int v2 = get(2 * v + 1, mid + 1, cr, mid + 1, r, (mod + v1) % 60);
    return v1 + v2;
  }
}
void update(int v, int cl, int cr, int ps, int value) {
  if (cl == cr) {
    for (int i = 0; i < 60; ++i) {
      if (i % value == 0) {
        t[v][i] = 2;
      } else {
        t[v][i] = 1;
      }
    }
    return;
  }
  int mid = (cl + cr) / 2;
  if (ps <= mid) {
    update(2 * v, cl, mid, ps, value);
  } else {
    update(2 * v + 1, mid + 1, cr, ps, value);
  }
  for (int i = 0; i < 60; ++i) {
    int v1 = t[2 * v][i];
    int v2 = t[2 * v + 1][(i + v1) % 60];
    t[v][i] = v1 + v2;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    update(1, 1, n, i, a[i]);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int x, y;
    string cur;
    cin >> cur;
    if (cur == "A") {
      cin >> x >> y;
      --y;
      cout << get(1, 1, n, x, y, 0) << endl;
    } else if (cur == "C") {
      cin >> x >> y;
      update(1, 1, n, x, y);
    }
  }
  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 maxN = 101000;
int t[4 * maxN][60];
int a[maxN];
int get(int v, int cl, int cr, int l, int r, int mod) {
  if (cl == l && cr == r) {
    return t[v][mod];
  }
  int mid = (cl + cr) / 2;
  if (r <= mid) {
    return get(2 * v, cl, mid, l, r, mod);
  } else if (l > mid) {
    return get(2 * v + 1, mid + 1, cr, l, r, mod);
  } else {
    int v1 = get(2 * v, cl, mid, l, mid, mod);
    int v2 = get(2 * v + 1, mid + 1, cr, mid + 1, r, (mod + v1) % 60);
    return v1 + v2;
  }
}
void update(int v, int cl, int cr, int ps, int value) {
  if (cl == cr) {
    for (int i = 0; i < 60; ++i) {
      if (i % value == 0) {
        t[v][i] = 2;
      } else {
        t[v][i] = 1;
      }
    }
    return;
  }
  int mid = (cl + cr) / 2;
  if (ps <= mid) {
    update(2 * v, cl, mid, ps, value);
  } else {
    update(2 * v + 1, mid + 1, cr, ps, value);
  }
  for (int i = 0; i < 60; ++i) {
    int v1 = t[2 * v][i];
    int v2 = t[2 * v + 1][(i + v1) % 60];
    t[v][i] = v1 + v2;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    update(1, 1, n, i, a[i]);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int x, y;
    string cur;
    cin >> cur;
    if (cur == "A") {
      cin >> x >> y;
      --y;
      cout << get(1, 1, n, x, y, 0) << endl;
    } else if (cur == "C") {
      cin >> x >> y;
      update(1, 1, n, x, y);
    }
  }
  return 0;
}

```