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 maxN = 100 * 1000 + 2;
const int mod = 1000 * 1000 * 1000 + 7;
int seg[60][4 * maxN];
int a[maxN];
void make(int ind, int s, int e) {
  if (e - s == 1) {
    for (int i = 0; i < 60; i++)
      if (i % a[s] == 0)
        seg[i][ind] = 2;
      else
        seg[i][ind] = 1;
    return;
  }
  int mid = (s + e) >> 1;
  make(2 * ind + 0, s, mid);
  make(2 * ind + 1, mid, e);
  for (int i = 0; i < 60; i++) {
    int tmp = (seg[i][2 * ind] + i) % 60;
    seg[i][ind] = seg[tmp][2 * ind + 1] + seg[i][2 * ind];
  }
}
void mof(int p, int val, int s, int e, int ind) {
  if (e - s == 1) {
    for (int i = 0; i < 60; i++)
      if (i % val == 0)
        seg[i][ind] = 2;
      else
        seg[i][ind] = 1;
    return;
  }
  int mid = (s + e) >> 1;
  if (p < mid)
    mof(p, val, s, mid, 2 * ind);
  else
    mof(p, val, mid, e, 2 * ind + 1);
  for (int i = 0; i < 60; i++) {
    int tmp = (seg[i][2 * ind] + i) % 60;
    seg[i][ind] = seg[tmp][2 * ind + 1] + seg[i][2 * ind];
  }
}
int get(int l, int r, int ti, int s, int e, int ind) {
  if (l >= e || r <= s) return ti;
  if (l <= s && e <= r) {
    return ti + seg[ti % 60][ind];
  }
  int mid = (s + e) >> 1;
  ti = get(l, r, ti, s, mid, 2 * ind + 0);
  ti = get(l, r, ti, mid, e, 2 * ind + 1);
  return ti;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  make(1, 0, n);
  int m;
  cin >> m;
  while (m--) {
    char c;
    int x, y;
    cin >> c >> x >> y;
    x--;
    if (c == 'A') {
      y--;
      cout << get(x, y, 0, 0, n, 1) << endl;
      continue;
    }
    mof(x, y, 0, n, 1);
  }
  return 0;
}
