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 + 5;
const int mod = 1e9 + 7;
const int P = 60;
int n, i, q, x, y;
int A[N];
char c;
class node {
 public:
  int A[P];
  int operator==(node x) {
    for (int i = 0; i < P; i++)
      if (A[i] != x.A[i]) return 0;
    return 1;
  }
};
node first[N * 4], e;
node merge(node x, node y) {
  if (x == e) return y;
  if (y == e) return x;
  node ans;
  for (int i = 0; i < P; i++) ans.A[i] = y.A[(i + x.A[i]) % 60] + x.A[i];
  return ans;
}
node query(int root, int bas, int son, int x, int y) {
  if (bas > y or son < x) return e;
  if (x <= bas and son <= y) return first[root];
  return merge(query((root + root), bas, ((bas + son) / 2), x, y),
               query((root + root + 1), ((bas + son) / 2) + 1, son, x, y));
}
node update(int root, int bas, int son, int x, int v) {
  if (bas > x or son < x) return first[root];
  if (bas == son) {
    for (int i = 0; i < P; i++) first[root].A[i] = (i % v) ? 1 : 2;
    return first[root];
  }
  return first[root] =
             merge(update((root + root), bas, ((bas + son) / 2), x, v),
                   update((root + root + 1), ((bas + son) / 2) + 1, son, x, v));
}
node init(int root, int bas, int son) {
  if (bas == son) {
    for (int i = 0; i < P; i++) first[root].A[i] = (i % A[bas]) ? 1 : 2;
    return first[root];
  }
  return first[root] =
             merge(init((root + root), bas, ((bas + son) / 2)),
                   init((root + root + 1), ((bas + son) / 2) + 1, son));
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", A + i);
  init(1, 1, n);
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf(" %c %d %d", &c, &x, &y);
    if (c == 'C')
      update(1, 1, n, x, y);
    else
      printf("%d\n", query(1, 1, n, x, y - 1).A[0]);
  }
}
