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 T>
inline T gcd(T a, T b) {
  return (b) == 0 ? (a) : gcd((b), ((a) % (b)));
}
template <class T>
inline T lcm(T a, T b) {
  return ((a) / gcd((a), (b)) * (b));
}
template <class T>
inline T BigMod(T Base, T power, T M = 1000000007) {
  if (power == 0) return 1;
  if (power & 1)
    return ((Base % M) * (BigMod(Base, power - 1, M) % M)) % M;
  else {
    T y = BigMod(Base, power / 2, M) % M;
    return (y * y) % M;
  }
}
template <class T>
inline T ModInv(T A, T M = 1000000007) {
  return BigMod(A, M - 2, M);
}
int fx[] = {-1, +0, +1, +0, +1, +1, -1, -1, +0};
int fy[] = {+0, -1, +0, +1, +1, -1, +1, -1, +0};
int day[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int arr[100005];
int tree[4 * 100005][61];
void update(int node, int b, int e, int indx) {
  if (b > indx or indx > e) return;
  if (b == e) {
    for (int i = 0; i < 60; i++) tree[node][i] = 1 + (i % arr[indx] == 0);
    return;
  }
  int mid = (b + e) >> 1;
  int left = 2 * node;
  int right = left + 1;
  update(left, b, mid, indx);
  update(right, mid + 1, e, indx);
  for (int i = 0; i < 60; i++) {
    tree[node][i] = tree[left][i] + tree[right][(tree[left][i] + i) % 60];
  }
}
int query(int node, int b, int e, int l, int r, int t) {
  if (b > r or l > e) return 0;
  if (b >= l && e <= r) return tree[node][t];
  int mid = (b + e) >> 1;
  int left = 2 * node;
  int right = left + 1;
  int p1 = query(left, b, mid, l, r, t);
  int p2 = query(right, mid + 1, e, l, r, (p1 + t) % 60);
  return p1 + p2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    update(1, 1, n, i);
  }
  int q;
  cin >> q;
  while (q--) {
    char c;
    int l, r;
    cin >> c >> l >> r;
    if (c == 'A') {
      cout << query(1, 1, n, l, r - 1, 0) << "\n";
    } else {
      arr[l] = r;
      update(1, 1, n, l);
    }
  }
  return 0;
}
