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 = (int)1e5, LCM = 60;
int A[N + 1], segtree[LCM][4 * N];
void init(int id, int from, int to) {
  if (from + 1 == to) {
    for (int i = 0; i < LCM; i++) segtree[i][id] = i % A[from] == 0 ? 2 : 1;
    return;
  }
  int middle = (from + to) / 2;
  init(2 * id + 1, from, middle);
  init(2 * id + 2, middle, to);
  for (int i = 0; i < LCM; i++)
    segtree[i][id] = segtree[i][2 * id + 1] +
                     segtree[(i + segtree[i][2 * id + 1]) % LCM][2 * id + 2];
}
void update(int id, int from, int to, int at) {
  if (from > at || to <= at) return;
  if (from + 1 == to) {
    for (int i = 0; i < LCM; i++) segtree[i][id] = i % A[from] == 0 ? 2 : 1;
    return;
  }
  int middle = (from + to) / 2;
  update(2 * id + 1, from, middle, at);
  update(2 * id + 2, middle, to, at);
  for (int i = 0; i < LCM; i++)
    segtree[i][id] = segtree[i][2 * id + 1] +
                     segtree[(i + segtree[i][2 * id + 1]) % LCM][2 * id + 2];
}
int query(int id, int from, int to, int left, int right, int shift) {
  if (from >= right || to <= left) return -1;
  if (from >= left && to <= right) {
    return shift + segtree[shift % LCM][id];
  }
  int middle = (from + to) / 2;
  if (right <= middle)
    return query(2 * id + 1, from, middle, left, right, shift);
  if (left >= middle) return query(2 * id + 2, middle, to, left, right, shift);
  int ret = query(2 * id + 1, from, middle, left, right, shift);
  return query(2 * id + 2, middle, to, left, right, ret);
}
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &A[i]);
  init(0, 0, n);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    char c;
    int x, y;
    cin >> c >> x >> y;
    if (c == 'A')
      printf("%d\n", query(0, 0, n, x - 1, y - 1, 0));
    else {
      A[--x] = y;
      update(0, 0, n, x);
    }
  }
}
int main() {
  solve();
  return 0;
}
