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 TAM = 1e5 + 10;
struct node {
  int val[61];
  long long calc;
  int from, to;
};
node ST[TAM * 4];
int arr[TAM];
void make_tree(int pos, int from, int to) {
  node &x = ST[pos];
  x.from = from;
  x.to = to;
  x.calc = 0ll;
  if (from == to) {
    return;
  }
  int mid = (from + to) / 2;
  make_tree((pos * 2), from, mid);
  make_tree((pos * 2 + 1), mid + 1, to);
}
void change(int pos, int dest, int val) {
  node &x = ST[pos];
  x.calc = 0ll;
  if (x.from == x.to && x.from == dest) {
    arr[x.from] = val;
    return;
  }
  if (dest <= ST[(pos * 2)].to)
    change((pos * 2), dest, val);
  else
    change((pos * 2 + 1), dest, val);
}
int calc(int pos, int from, int to, int st) {
  node &x = ST[pos];
  if (x.from == x.to) return st + 1 + ((st % arr[x.from]) == 0);
  if (x.from == from && x.to == to && (x.calc & (1ll << st)) != 0)
    return x.val[st];
  if (to <= ST[(pos * 2)].to) return calc((pos * 2), from, to, st);
  if (from >= ST[(pos * 2 + 1)].from) return calc((pos * 2 + 1), from, to, st);
  int tot = 0;
  int L = calc((pos * 2), from, ST[(pos * 2)].to, st);
  tot += L;
  tot -= (L % 60);
  int R = calc((pos * 2 + 1), ST[(pos * 2 + 1)].from, to, (L % 60));
  tot += R;
  if (x.from == from && x.to == to) {
    x.calc |= (1ll << st);
    x.val[st] = tot;
  }
  return tot;
}
int main() {
  char type;
  int n, nQ, x, y;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  make_tree(1, 1, n);
  scanf("%d", &nQ);
  for (int i = 0; i < nQ; i++) {
    scanf(" %c %d %d", &type, &x, &y);
    if (type == 'C')
      change(1, x, y);
    else
      printf("%d\n", calc(1, x, y - 1, 0));
  }
  return 0;
}
