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 = 1e5 + 50, B = 60, MOD = 0;
int seg[4 * N][B + 5], n, a[N], q;
void build(int l = 0, int r = n, int id = 1) {
  if (l + 1 == r) {
    for (int i = 0; i < B; i++) {
      if (i % a[l] == 0)
        seg[id][i] = 2;
      else
        seg[id][i] = 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, id * 2);
  build(mid, r, id * 2 + 1);
  for (int i = 0; i < B; i++)
    seg[id][i] = seg[id * 2][i] + seg[id * 2 + 1][(seg[id * 2][i] + i) % B];
  return;
}
void upd(int ql, int qr, int val, int l = 0, int r = n, int id = 1) {
  if (l >= qr || r <= ql) return;
  if (ql <= l && qr >= r) {
    a[ql] = val;
    for (int i = 0; i < B; i++) {
      if (i % a[ql] == 0)
        seg[id][i] = 2;
      else
        seg[id][i] = 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  upd(ql, qr, val, l, mid, id * 2);
  upd(ql, qr, val, mid, r, id * 2 + 1);
  for (int i = 0; i <= B; i++)
    seg[id][i] = seg[id * 2][i] + seg[id * 2 + 1][(seg[id * 2][i] + i) % B];
  return;
}
int get(int ql, int qr, int t = 0, int l = 0, int r = n, int id = 1) {
  if (l >= qr || ql >= r) return 0;
  if (ql <= l && qr >= r) return seg[id][t];
  int mid = (l + r) / 2;
  int tmp = 0;
  if (ql < mid) tmp += get(ql, qr, t, l, mid, id * 2);
  t = (t + tmp) % B;
  if (qr > mid) tmp += get(ql, qr, t, mid, r, id * 2 + 1);
  return tmp;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  build();
  cin >> q;
  for (int i = 0; i < q; i++) {
    char c;
    int b, d;
    cin >> c >> b >> d;
    if (c == 'C')
      upd(b - 1, b, d);
    else
      cout << get(b - 1, d - 1) << '\n';
  }
  return 0;
}
