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 inf = 1e9 + 5;
const long long linf = 1e18 + 5;
const int N = 1e5 + 5;
const int LCM = 60;
int n, m, x, y, a[N], kd[4 * N][LCM];
char c;
void init(int root, int bas, int son) {
  if (bas == son) {
    for (int i = 0; i <= LCM - 1; i++) kd[root][i] = 1 + (i % a[bas] == 0);
    return;
  }
  init((root << 1), bas, ((bas + son) >> 1));
  init(((root << 1) | 1), ((bas + son) >> 1) + 1, son);
  for (int i = 0; i <= LCM - 1; i++)
    kd[root][i] = kd[(root << 1)][i] +
                  kd[((root << 1) | 1)][(i + kd[(root << 1)][i]) % 60];
}
int query(int root, int bas, int son, int x, int y, int time) {
  if (son < x or y < bas) return 0;
  if (x <= bas and son <= y) return kd[root][time % 60];
  int start = time;
  time += query((root << 1), bas, ((bas + son) >> 1), x, y, time);
  time += query(((root << 1) | 1), ((bas + son) >> 1) + 1, son, x, y, time);
  return time - start;
}
void update(int root, int bas, int son, int x, int val) {
  if (son < x or x < bas) return;
  if (x <= bas and son <= x) {
    a[x] = val;
    for (int i = 0; i <= LCM - 1; i++) kd[root][i] = 1 + (i % val == 0);
    return;
  }
  update((root << 1), bas, ((bas + son) >> 1), x, val);
  update(((root << 1) | 1), ((bas + son) >> 1) + 1, son, x, val);
  for (int i = 0; i <= LCM - 1; i++)
    kd[root][i] = kd[(root << 1)][i] +
                  kd[((root << 1) | 1)][(i + kd[(root << 1)][i]) % 60];
}
int main() {
  ios ::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  init(1, 1, n);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> c >> x >> y;
    if (c == 'A')
      cout << query(1, 1, n, x, y - 1, 0) << '\n';
    else
      update(1, 1, n, x, y);
  }
  return 0;
}
