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>
std::vector<int> merge(std::vector<int> a, std::vector<int> b) {
  std::vector<int> res(60);
  for (int i = 0; i < 60; ++i) {
    res[i] = a[i] + b[(i + a[i]) % 60];
  }
  return res;
}
const int shift = 1 << 17;
std::vector<std::vector<int> > tree(shift << 1, std::vector<int>(60));
std::vector<int> get(int l, int r) {
  if (l >= r) return std::vector<int>(60, 0);
  if (l % 2) {
    return merge(tree[l], get(l + 1, r));
  }
  if (r % 2) {
    --r;
    return merge(get(l, r), tree[r]);
  }
  return get(l / 2, r / 2);
}
class A {
 public:
  void solve(std::istream& in, std::ostream& out) {
    int n;
    in >> n;
    std::vector<int> a(n);
    for (int i = 0; i < n; ++i) {
      in >> a[i];
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < 60; ++j) {
        if (j % a[i] == 0)
          tree[i + shift][j] = 2;
        else
          tree[i + shift][j] = 1;
      }
    }
    for (int i = shift - 1; i; --i) {
      tree[i] = merge(tree[2 * i], tree[2 * i + 1]);
    }
    int m;
    in >> m;
    for (int it = 0; it < m; ++it) {
      char c;
      int x, y;
      in >> c >> x >> y;
      if (c == 'C') {
        --x;
        a[x] = y;
        int i = x;
        for (int j = 0; j < 60; ++j) {
          if (j % a[i] == 0)
            tree[i + shift][j] = 2;
          else
            tree[i + shift][j] = 1;
        }
        i += shift;
        i /= 2;
        while (i) {
          tree[i] = merge(tree[2 * i], tree[2 * i + 1]);
          i /= 2;
        }
      } else {
        --x, --y;
        out << get(x + shift, y + shift)[0] << "\n";
      }
    }
  }
};
int main() {
  std::cin.sync_with_stdio(false);
  std::cin.tie(0);
  std::cout << std::fixed;
  std::cout.precision(20);
  A solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
