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>
const int MOD = 60, Cv = 1 << 17, C = 20;
int val[2 * Cv][MOD];
int palingnesia[C], ln = 0;
void reformulate_all(int a) {
  for (int j = 0; j < MOD; j++)
    val[a][j] = val[a * 2][j] + val[a * 2 + 1][(j + val[a * 2][j]) % MOD];
}
int query(int l, int r) {
  int cur_res = val[l + Cv][0];
  ln = 0;
  if (l != r) palingnesia[ln] = r + Cv, ln++;
  for (l += Cv, r += Cv; l > 0; l /= 2, r /= 2) {
    if (l % 2 == 0 && r - l > 1) cur_res += val[l + 1][cur_res % MOD];
    if (r % 2 == 1 && r - l > 1) palingnesia[ln] = r - 1, ln++;
  }
  for (ln--; ln >= 0; ln--) cur_res += val[palingnesia[ln]][cur_res % MOD];
  return cur_res;
}
void insert(int p, int v) {
  int j;
  p += Cv;
  for (j = 0; j < MOD; j++) val[p][j] = ((j % v == 0) ? 2 : 1);
  for (p /= 2; p > 0; p /= 2) {
    reformulate_all(p);
  }
}
int main() {
  int n, a, i, j, b, q;
  char c, tmp;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a);
    for (j = 0; j < MOD; j++) val[Cv + i][j] = ((j % a == 0) ? 2 : 1);
  }
  for (i = Cv - 1; i >= 1; i--) {
    reformulate_all(i);
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%c%c %d %d", &tmp, &c, &a, &b);
    if (c == 'A')
      printf("%d\n", query(a - 1, b - 2));
    else
      insert(a - 1, b);
  }
  return 0;
}
