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 MAXN = 100002;
int n, a[MAXN];
int t[4 * MAXN][60];
void build(int v, int l, int r) {
  if (l + 1 == r) {
    for (int i = (0); i < (60); i++) t[v][i] = (i % a[l]) ? 1 : 2;
    return;
  }
  int m = (l + r) / 2;
  build(2 * v, l, m);
  build(2 * v + 1, m, r);
  for (int i = (0); i < (60); i++)
    t[v][i] = t[2 * v][i] + t[2 * v + 1][(i + t[2 * v][i]) % 60];
}
void upd(int p, int v, int l, int r) {
  if (l + 1 == r) {
    for (int i = (0); i < (60); i++) t[v][i] = (i % a[l]) ? 1 : 2;
    return;
  }
  int m = (l + r) / 2;
  if (p < m)
    upd(p, 2 * v, l, m);
  else
    upd(p, 2 * v + 1, m, r);
  for (int i = (0); i < (60); i++) {
    t[v][i] = t[2 * v][i] + t[2 * v + 1][(i + t[2 * v][i]) % 60];
  }
}
int get(int i, int j, int v, int l, int r, int ct = 0) {
  if (i == l && j == r) return t[v][ct % 60];
  int m = (l + r) / 2;
  if (j <= m)
    return get(i, j, 2 * v, l, m, ct);
  else if (i >= m)
    return get(i, j, 2 * v + 1, m, r, ct);
  int t1 = get(i, m, 2 * v, l, m, ct);
  ct += t1;
  int t2 = get(m, j, 2 * v + 1, m, r, ct);
  return t1 + t2;
}
int main() {
  while (cin >> n) {
    for (int i = (0); i < (n); i++) cin >> a[i];
    build(1, 0, n);
    int q;
    cin >> q;
    while (q--) {
      char cmd;
      int x, y;
      cin >> cmd >> x >> y;
      if (cmd == 'A') {
        cout << get(x - 1, y - 1, 1, 0, n) << endl;
      } else {
        a[x - 1] = y;
        upd(x - 1, 1, 0, n);
      }
    }
  }
  return 0;
}
