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;
int n;
int tree[4 * 100010][61];
void leafup(int ind, int a) {
  for (int i = 0; i < 60; ++i) {
    if (i % a == 0)
      tree[ind][i] = 2;
    else
      tree[ind][i] = 1;
  }
}
void nodeup(int ind) {
  for (int i = 0; i < 60; ++i) {
    int x = tree[2 * ind][i];
    int j = (i + x) % 60;
    tree[ind][i] = x + tree[2 * ind + 1][j];
  }
}
void build(int ind, int lo, int hi) {
  if (lo == hi) {
    int a;
    cin >> a;
    leafup(ind, a);
    return;
  }
  int mid = (lo + hi) >> 1;
  build(2 * ind, lo, mid);
  build(2 * ind + 1, mid + 1, hi);
  nodeup(ind);
}
void update(int ind, int lo, int hi, int arrind, int val) {
  if (lo == hi) {
    leafup(ind, val);
    return;
  }
  int mid = (lo + hi) >> 1;
  if (arrind <= mid)
    update(2 * ind, lo, mid, arrind, val);
  else
    update(2 * ind + 1, mid + 1, hi, arrind, val);
  nodeup(ind);
}
int query(int ind, int lo, int hi, int x, int y, int time) {
  if (lo > y or hi < x) return 0;
  if (lo >= x and hi <= y) return tree[ind][time];
  int mid = (lo + hi) >> 1;
  int q1 = query(2 * ind, lo, mid, x, y, time);
  int t2 = (q1 + time) % 60;
  return q1 + query(2 * ind + 1, mid + 1, hi, x, y, t2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  build(1, 0, n - 1);
  int q;
  cin >> q;
  while (q--) {
    char c;
    int x, y;
    cin >> c >> x >> y;
    if (c == 'A') {
      cout << query(1, 0, n - 1, x - 1, y - 2, 0) << "\n";
    } else
      update(1, 0, n - 1, x - 1, y);
  }
  return 0;
}
