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 seg_tree[100100 * 2][60];
int a[100100];
void upd(int x, int n) {
  if (x >= 2 * n) return;
  int b = (x & -x) >> 1, l = x - b, r = x + b;
  int i, tmp;
  for (i = 0; i < 60; i++) {
    tmp = seg_tree[l][i];
    seg_tree[x][i] = tmp + (r < 2 * n ? seg_tree[r][(i + tmp) % 60] : 0);
  }
}
int main() {
  int i, j, k, l, n, q, tmp, x, y;
  char op[5];
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    for (j = 0; j < 60; j++) {
      if (j % a[i] == 0) {
        seg_tree[i * 2 - 1][j] = 2;
      } else
        seg_tree[i * 2 - 1][j] = 1;
    }
  }
  for (i = 2; i < n * 2; i <<= 1) {
    for (j = i; j < n * 2; j += (i << 1)) {
      upd(j, n);
    }
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%s%d%d", op, &x, &y);
    if (op[0] == 'A') {
      y = 2 * y - 3;
      x = 2 * x - 1;
      i = 0;
      while (x <= y) {
        j = x + (x & -x);
        if ((j & -j) != ((x & -x) << 1) || j + (j & -j) > y) {
          i += seg_tree[x][i % 60];
          x = j + 1;
        } else
          x = j;
      }
      printf("%d\n", i);
    } else {
      a[x] = y;
      for (i = 0; i < 60; i++) seg_tree[2 * x - 1][i] = 1 + (i % y == 0);
      for (i = 2 * x - 1; i < 2 * n;) {
        k = i - (i & -i);
        l = i + (i & -i);
        if ((k & -k) == ((i & -i) << 1)) {
          upd(k, n);
          i = k;
        } else {
          upd(l, n);
          i = l;
        }
      }
    }
  }
  return 0;
}
