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;
template <class P, class Q>
inline void smin(P &a, Q b) {
  if (b < a) a = b;
}
template <class P, class Q>
inline void smax(P &a, Q b) {
  if (a < b) a = b;
}
const int maxn = 100000 + 1000;
const int lcm = 60;
int n, q;
int a[maxn];
int p[4 * maxn][lcm];
void update(int x) {
  for (int m = 0, _n = (int)(lcm); m < _n; m++) {
    int t = p[2 * x + 1][m];
    p[x][m] = t + p[2 * x + 2][(t + m) % lcm];
  }
}
void make(int x, int xl, int xr) {
  if (xr - xl == 1) {
    for (int m = 0, _n = (int)(lcm); m < _n; m++) p[x][m] = (m % a[xl] ? 1 : 2);
    return;
  }
  int xm = (xl + xr) / 2;
  make(2 * x + 1, xl, xm);
  make(2 * x + 2, xm, xr);
  update(x);
}
void change(int x, int xl, int xr, int qpos) {
  if (qpos < xl || xr <= qpos) return;
  if (xr - xl == 1) {
    for (int m = 0, _n = (int)(lcm); m < _n; m++) p[x][m] = (m % a[xl] ? 1 : 2);
    return;
  }
  int xm = (xl + xr) / 2;
  change(2 * x + 1, xl, xm, qpos);
  change(2 * x + 2, xm, xr, qpos);
  update(x);
}
int ask(int x, int xl, int xr, int ql, int qr, int ct = 0) {
  if (qr <= xl || xr <= ql) return ct;
  if (ql <= xl && xr <= qr) {
    ct += p[x][ct % lcm];
    return ct;
  }
  int xm = (xl + xr) / 2;
  ct = ask(2 * x + 1, xl, xm, ql, qr, ct);
  ct = ask(2 * x + 2, xm, xr, ql, qr, ct);
  return ct;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0, _n = (int)(n); i < _n; i++) cin >> a[i];
  make(0, 0, n);
  cin >> q;
  for (int z = 0, _n = (int)(q); z < _n; z++) {
    char tp;
    cin >> tp;
    int x, y;
    cin >> x >> y;
    if (tp == 'A') {
      x--, y--;
      cout << ask(0, 0, n, x, y) << endl;
    } else {
      x--;
      a[x] = y;
      change(0, 0, n, x);
    }
  }
  return 0;
}
