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;
inline long long rd() {
  long long _x = 0;
  int _ch = getchar(), _f = 1;
  for (; !isdigit(_ch) && (_ch != '-') && (_ch != EOF); _ch = getchar())
    ;
  if (_ch == '-') {
    _f = 0;
    _ch = getchar();
  }
  for (; isdigit(_ch); _ch = getchar()) _x = _x * 10 + _ch - '0';
  return _f ? _x : -_x;
}
void write(long long _x) {
  if (_x >= 10)
    write(_x / 10), putchar(_x % 10 + '0');
  else
    putchar(_x + '0');
}
inline void wrt(long long _x, char _p) {
  if (_x < 0) putchar('-'), _x = -_x;
  write(_x);
  if (_p) putchar(_p);
}
int n, Q;
int a[100005];
struct node {
  int r[60];
} zero;
struct Tree {
  node tr[100005 << 2];
  node merge(node Ls, node Rs) {
    node ans;
    for (int i = int(0); i <= (int)(59); i++) {
      ans.r[i] = Ls.r[i] + Rs.r[(Ls.r[i] + i) % 60];
    }
    return ans;
  }
  void update(int pos, int l, int r, int w, int v) {
    if (l == r) {
      for (int i = int(0); i <= (int)(59); i++)
        tr[pos].r[i] = 1 + ((i % v) == 0);
      return;
    }
    int mid = (l + r) >> 1;
    if (w <= mid)
      update(pos << 1, l, mid, w, v);
    else
      update(pos << 1 | 1, mid + 1, r, w, v);
    tr[pos] = merge(tr[pos << 1], tr[pos << 1 | 1]);
  }
  node query(int pos, int l, int r, int ql, int qr) {
    if (qr < l || r < ql) return zero;
    if (ql <= l && r <= qr) return tr[pos];
    int mid = (l + r) >> 1;
    return merge(query(pos << 1, l, mid, ql, qr),
                 query(pos << 1 | 1, mid + 1, r, ql, qr));
  }
  void build(int pos, int l, int r) {
    if (l == r) {
      for (int i = int(0); i <= (int)(59); i++)
        tr[pos].r[i] = 1 + ((i % a[l]) == 0);
      return;
    }
    int mid = (l + r) >> 1;
    build(pos << 1, l, mid), build(pos << 1 | 1, mid + 1, r);
    tr[pos] = merge(tr[pos << 1], tr[pos << 1 | 1]);
  }
} tr;
int main() {
  n = rd();
  for (int i = int(1); i <= (int)(n); i++) a[i] = rd();
  Q = rd();
  tr.build(1, 1, n);
  for (int i = int(1); i <= (int)(Q); i++) {
    char ch = getchar();
    while (ch != 'A' && ch != 'C') ch = getchar();
    int x = rd(), y = rd();
    if (ch == 'A')
      wrt(x == y ? 0 : tr.query(1, 1, n, x, y - 1).r[0], '\n');
    else
      tr.update(1, 1, n, x, y);
  }
  return 0;
}
