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 N = 110000;
const int B = 60;
int n, q, mc;
int aint[4 * N][B];
void update(int nod, int pozx, int pozy, int poz, int val) {
  int i;
  if (pozx == pozy) {
    for (i = 0; i < B; ++i) {
      aint[nod][i] = 1;
      if (i % val == 0) aint[nod][i]++;
    }
    return;
  }
  int mid = (pozx + pozy) / 2;
  if (mid >= poz)
    update(2 * nod, pozx, mid, poz, val);
  else
    update(2 * nod + 1, mid + 1, pozy, poz, val);
  for (i = 0; i < B; ++i)
    aint[nod][i] =
        aint[2 * nod][i] + aint[2 * nod + 1][(i + aint[2 * nod][i]) % B];
}
int query(int nod, int pozx, int pozy, int poz1, int poz2) {
  if (poz1 <= pozx && pozy <= poz2) {
    int rez = aint[nod][mc];
    mc = (mc + aint[nod][mc]) % B;
    return rez;
  }
  int mid = (pozx + pozy) / 2, rez = 0;
  if (mid >= poz1) rez = query(2 * nod, pozx, mid, poz1, poz2);
  if (mid < poz2) rez += query(2 * nod + 1, mid + 1, pozy, poz1, poz2);
  return rez;
}
int main() {
  int i;
  cin >> n;
  for (i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    update(1, 1, n, i, x);
  }
  cin >> q;
  for (i = 1; i <= q; ++i) {
    char a;
    int p1, p2;
    cin >> a >> p1 >> p2;
    if (a == 'A') {
      mc = 0;
      cout << query(1, 1, n, p1, p2 - 1) << "\n";
    } else
      update(1, 1, n, p1, p2);
  }
  return 0;
}
