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>
int n;
int IT[262144][60];
void UDT(int a) {
  while (a != 1) {
    a >>= 1;
    for (int x = 0; x < 60; x++) {
      IT[a][x] = IT[a * 2][x] + IT[a * 2 + 1][(x + IT[a * 2][x]) % 60];
    }
  }
}
void Do(int a, int b) {
  a += 131072;
  int i;
  for (i = 0; i < 60; i++) {
    if (i % b == 0)
      IT[a][i] = 2;
    else
      IT[a][i] = 1;
  }
  UDT(a);
}
int Sum(int a, int b) {
  a += 131072, b += 131072;
  int P[20], cnt = 0, res = 0, i;
  while (a <= b) {
    if (a & 1) {
      res += IT[a][res % 60];
    }
    if (!(b & 1)) P[cnt++] = b;
    a = (a + 1) >> 1, b = (b - 1) >> 1;
  }
  for (i = cnt - 1; i >= 0; i--) {
    res += IT[P[i]][res % 60];
  }
  return res;
}
int main() {
  int i, Q, a, b;
  char p[2];
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a);
    Do(i, a);
  }
  scanf("%d", &Q);
  while (Q--) {
    scanf("%s", p);
    scanf("%d%d", &a, &b);
    if (p[0] == 'C') {
      Do(a, b);
    } else
      printf("%d\n", Sum(a, b - 1));
  }
}
