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 int N = 1e5 + 10;
int data[4 * N][60], a[N], n, q;
void change(int node, int st, int en, int p, int val) {
  if (en - st < 2) {
    for (int i = 0; i < 60; ++i) {
      data[node][i] = 1;
      if (i % val == 0) data[node][i]++;
    }
    return;
  }
  int mid = (st + en) / 2;
  if (p < mid)
    change(node * 2 + 0, st, mid, p, val);
  else
    change(node * 2 + 1, mid, en, p, val);
  int help;
  for (int i = 0; i < 60; ++i) {
    help = data[node * 2][i];
    help += i;
    help %= 60;
    data[node][i] = data[node * 2][i] + data[node * 2 + 1][help];
  }
}
int get(int node, int st, int en, int x, int l, int r) {
  if (en <= l or r <= st) return 0;
  if (l <= st and en <= r) return data[node][x];
  int mid = (st + en) / 2;
  int help1 = get(node * 2 + 0, st, mid, x, l, r);
  x += help1;
  x %= 60;
  int help2 = get(node * 2 + 1, mid, en, x, l, r);
  return help1 + help2;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    change(1, 0, n, i, a[i]);
  }
  cin >> q;
  char type;
  int l, r, val;
  while (q--) {
    cin >> type;
    if (type == 'A') {
      scanf("%d%d", &l, &r);
      l--;
      r--;
      cout << get(1, 0, n, 0, l, r) << endl;
    }
    if (type == 'C') {
      scanf("%d%d", &l, &val);
      l--;
      change(1, 0, n, l, val);
    }
  }
}
