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 SEGW = 1 << 17;
int N, Q, prd[SEGW + 5];
int seg[2 * SEGW + 5][60];
void build(int loc = 1, int x = 0, int y = SEGW - 1) {
  if (x == y) {
    for (int e = 0; e < 60; e++)
      if (e % prd[x])
        seg[loc][e] = 1;
      else
        seg[loc][e] = 2;
    return;
  }
  build(2 * loc, x, x + (y - x + 1) / 2 - 1);
  build(2 * loc + 1, x + (y - x + 1) / 2, y);
  for (int e = 0; e < 60; e++) {
    int fe = seg[2 * loc][e];
    int se = seg[2 * loc + 1][(e + fe) % 60];
    seg[loc][e] = fe + se;
  }
}
void upd(int i, int v, int loc = 1, int x = 0, int y = SEGW - 1) {
  if (x > i || y < i) return;
  if (x == y) {
    prd[x] = v;
    for (int e = 0; e < 60; e++)
      if (e % prd[x])
        seg[loc][e] = 1;
      else
        seg[loc][e] = 2;
    return;
  }
  upd(i, v, 2 * loc, x, x + (y - x + 1) / 2 - 1);
  upd(i, v, 2 * loc + 1, x + (y - x + 1) / 2, y);
  for (int e = 0; e < 60; e++) {
    int fe = seg[2 * loc][e];
    int se = seg[2 * loc + 1][(e + fe) % 60];
    seg[loc][e] = fe + se;
  }
}
int que(int e, int l, int r, int loc = 1, int x = 0, int y = SEGW - 1) {
  if (r < x || l > y) return 0;
  if (l <= x && y <= r) return seg[loc][e % 60];
  int fe = que(e, l, r, 2 * loc, x, x + (y - x + 1) / 2 - 1);
  int se = que(e + fe, l, r, 2 * loc + 1, x + (y - x + 1) / 2, y);
  return fe + se;
}
int main() {
  cin >> N;
  for (int i = 0; i < SEGW; i++) prd[i] = 6;
  for (int i = 0; i < N; i++) cin >> prd[i];
  build();
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    char t;
    cin >> t;
    if (t == 'C') {
      int x, v;
      cin >> x >> v;
      upd(x - 1, v);
    } else {
      int x, y;
      cin >> x >> y;
      cout << que(0, x - 1, y - 2) << "\n";
    }
  }
  cout.flush();
  return 0;
}
