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 double PI = 3.14159265358979323846;
const long long MODUL = 1000000007;
const long long MAXINT = 1000000010;
const long long MAXLL = 1000000000000000000;
template <typename T>
inline T MIN(T lef, T ri) {
  return (lef < ri ? lef : ri);
}
template <typename T>
inline T MAX(T lef, T ri) {
  return (lef > ri ? lef : ri);
}
template <typename T>
inline T MOD(T val) {
  return (val > 0 ? val : -val);
}
#pragma comment(linker, "/STACK:167772160")
int str[500000][60];
void ins(int l, int r, int now, int val, int pos) {
  if (r == l) {
    for (int i(0); i < 60; ++i)
      if (i % val)
        str[now][i] = 1;
      else
        str[now][i] = 2;
    return;
  }
  int mid = (r + l) / 2;
  if (mid < pos)
    ins(mid + 1, r, now * 2 + 1, val, pos);
  else
    ins(l, mid, now * 2, val, pos);
  for (int i(0); i < 60; ++i) {
    int t = str[now * 2][i];
    str[now][i] = t;
    str[now][i] += str[now * 2 + 1][(t + i) % 60];
  }
}
int get(int l, int r, int now, int L, int R, int T) {
  if (L > R) return 0;
  if (l == L && r == R) return str[now][T];
  int del = get(l, (l + r) / 2, now * 2, L, MIN(R, (l + r) / 2), T);
  return del + get((l + r) / 2 + 1, r, now * 2 + 1, MAX(L, (l + r) / 2 + 1), R,
                   (T + del) % 60);
}
int main() {
  int n, q, x, y;
  char c;
  cin >> n;
  for (int i(0); i < n; ++i) {
    cin >> x;
    ins(0, n - 1, 1, x, i);
  }
  cin >> q;
  while (q--) {
    cin >> c >> x >> y;
    if (c == 'C')
      ins(0, n - 1, 1, y, x - 1);
    else
      cout << get(0, n - 1, 1, x - 1, y - 2, 0) << '\n';
  }
  return 0;
}
