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;
int N;
int A[100000];
char c;
int Q;
int X, Y;
struct DAT {
  int c[60];
};
DAT mer(DAT d1, DAT d2) {
  if (d1.c[0] == -1) return d2;
  if (d2.c[0] == -1) return d1;
  DAT res;
  for (int(i) = 0; (i) < (60); (i)++)
    res.c[i] = d2.c[(i + d1.c[i]) % 60] + d1.c[i];
  return res;
}
const int n = 1 << 17;
struct SegTree {
  DAT d[n * 2 - 1];
  void reinit(int k, int c) {
    k += n - 1;
    memset((d[k].c), (0), sizeof(d[k].c));
    for (int(i) = 0; (i) < (60); (i)++) d[k].c[i] = 1 + (i % c == 0);
    while (k) k = (k - 1) / 2, d[k] = mer(d[k * 2 + 1], d[k * 2 + 2]);
  }
  DAT query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) {
      DAT res;
      res.c[0] = -1;
      return res;
    }
    if (a <= l && r <= b) return d[k];
    return mer(query(a, b, k * 2 + 1, l, (l + r) / 2),
               query(a, b, k * 2 + 2, (l + r) / 2, r));
  }
} seg;
int main() {
  scanf("%d", &N);
  for (int(i) = 0; (i) < (N); (i)++) scanf("%d", &A[i]);
  for (int(i) = 0; (i) < (N); (i)++) seg.reinit(i, A[i]);
  scanf("%d", &Q);
  while (Q--) {
    scanf(" %c%d%d", &c, &X, &Y);
    if (c == 'A') {
      printf("%d\n", seg.query(X - 1, Y - 1, 0, 0, n).c[0]);
    } else
      seg.reinit(X - 1, Y);
  }
}
