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;
struct Node {
  Node *l, *r;
  int a, b, mt, dt[60];
  void recompute() {
    if (b - a > 1) {
      mt = l->mt;
      for (auto t = (0); t < 60; t++) dt[t] = r->query(l->query(t)) - t;
    } else
      for (auto t = (0); t < 60; t++) dt[t] = 1 + (t % mt == 0);
  }
  void update(int i, int _mt) {
    if (i < a || b <= i) return;
    if (b - a == 1)
      mt = _mt;
    else
      l->update(i, _mt), r->update(i, _mt);
    recompute();
  }
  int query(int t, int i = 0, int j = 100005) {
    if (j <= a || b <= i) return t;
    if (i <= a && b <= j) return t + dt[t % 60];
    return r->query(l->query(t, i, j), i, j);
  }
  Node* init(int i, int j) {
    a = i, b = j, l = r = nullptr;
    int c = (a + b) / 2;
    Node* ans = this + 1;
    if (b - a == 1)
      cin >> mt;
    else
      ans = (r = (l = ans)->init(a, c))->init(c, b);
    recompute();
    return ans;
  }
};
Node buf[2 * 100005], *tree = buf;
int main() {
  ios_base::sync_with_stdio(false);
  int n, q;
  cin >> n;
  tree->init(0, n);
  cin >> q;
  string cmd;
  int x, y;
  for (auto i = (0); i < q; i++) {
    cin >> cmd >> x >> y;
    x--;
    if (cmd[0] == 'A')
      cout << tree->query(0, x, y - 1) << endl;
    else
      tree->update(x, y);
  }
  return 0;
}
