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
#include <bits/stdc++.h>
using namespace std;
const int M = 2500;
const int T = 60;
int n, m;
int a[100005];
int t[T][M];
void change(int x, int y) {
  int i = (x / m) * m;
  a[x] = y;
  int f = x - i;
  for (int(j) = (0); (j) < (T); ++(j)) {
    int cur = j;
    for (int k = 0; k < m && i + k < n; ++k) {
      while (cur % a[i + k] == 0) cur++;
      cur++;
    }
    t[j][i / m] = cur;
  }
}
int calc(int x, int y) {
  int res = 0;
  while (x % m && x < y) {
    while (res % a[x] == 0) res++;
    x++;
    res++;
  }
  while (x + m <= y) {
    int d = res % T;
    res += t[d][x / m] - d;
    x += m;
  }
  while (x < y) {
    while (res % a[x] == 0) res++;
    x++;
    res++;
  }
  return res;
}
int main() {
  cin >> n;
  for (int(i) = (0); (i) < (n); ++(i)) cin >> a[i];
  m = sqrt(n / 60.0);
  m = max(m, 1);
  for (int i = 0; i < n; i += m) {
    for (int(j) = (0); (j) < (T); ++(j)) {
      int cur = j;
      for (int k = 0; k < m && i + k < n; ++k) {
        while (cur % a[i + k] == 0) cur++;
        cur++;
      }
      t[j][i / m] = cur;
    }
  }
  int q;
  cin >> q;
  while (q--) {
    char c;
    int x, y;
    cin >> c >> x >> y;
    x--;
    if (c == 'A') {
      cout << calc(x, y - 1) << endl;
    } else {
      change(x, y);
    }
  }
  return 0;
}
