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;
namespace IO {
void read(int &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while ('0' > c || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  x *= f;
}
void write(int x) {
  if (x < 10)
    putchar(x + '0');
  else {
    write(x / 10);
    putchar(x % 10 + '0');
  }
}
}  // namespace IO
namespace lst {
int a[100005];
struct node {
  int l, r;
  int val[65];
} tre[400005];
void update(int k) {
  for (int i = 0; i < 60; i++)
    tre[k].val[i] =
        tre[k << 1].val[i] + tre[k << 1 | 1].val[(i + tre[k << 1].val[i]) % 60];
}
void build(int l, int r, int k) {
  tre[k].l = l;
  tre[k].r = r;
  if (l == r) {
    cin >> a[l];
    for (int i = 0; i < 60; i++) {
      if (i % a[l] == 0)
        tre[k].val[i] = 2;
      else
        tre[k].val[i] = 1;
    }
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, k << 1);
  build(mid + 1, r, k << 1 | 1);
  update(k);
}
void change(int pos, int val, int k = 1) {
  if (tre[k].l > pos || pos > tre[k].r) return;
  if (tre[k].l == tre[k].r) {
    a[tre[k].l] = val;
    for (int i = 0; i < 60; i++) {
      if (i % a[tre[k].l] == 0)
        tre[k].val[i] = 2;
      else
        tre[k].val[i] = 1;
    }
    return;
  }
  change(pos, val, k << 1);
  change(pos, val, k << 1 | 1);
  update(k);
}
void ask(int l, int r, int &now, int k = 1) {
  if (tre[k].l > r || l > tre[k].r) return;
  if (l <= tre[k].l && tre[k].r <= r) {
    now = now + tre[k].val[now % 60];
    return;
  }
  if (r <= tre[k << 1].r)
    ask(l, r, now, k << 1);
  else if (tre[k << 1 | 1].l <= l)
    ask(l, r, now, k << 1 | 1);
  else {
    ask(l, r, now, k << 1);
    ask(l, r, now, k << 1 | 1);
  }
}
}  // namespace lst
using namespace IO;
using namespace lst;
int n, m;
char opt;
int l, r;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  build(1, n, 1);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> opt >> l >> r;
    if (opt == 'C')
      change(l, r);
    else {
      int now = 0;
      if (l > r) swap(l, r);
      r--;
      if (l > r) {
        cout << "0\n";
        continue;
      }
      ask(l, r, now);
      cout << now << '\n';
    }
  }
  return 0;
}
