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;
static const int GCD = 60;
static const int MAXN = 100005;
int n;
int ai[MAXN];
int tim[MAXN * 4][GCD + 1];
void build(int start, int end, int node) {
  if (start == end) {
    for (int T = 0; T < GCD; T++) {
      if (T % ai[start] == 0)
        tim[node][T] = 2;
      else
        tim[node][T] = 1;
    }
    return;
  }
  build(start, (start + end) / 2, 2 * node);
  build((start + end) / 2 + 1, end, 2 * node + 1);
  for (int T = 0; T < GCD; T++) {
    tim[node][T] =
        tim[2 * node][T] + tim[2 * node + 1][(T + tim[2 * node][T]) % GCD];
  }
}
int query(int start, int end, int i, int j, int node, int T = 0) {
  if (start > j || end < i) return 0;
  if (start >= i && end <= j) return tim[node][T];
  int X = query(start, (start + end) / 2, i, j, 2 * node, T);
  int Y = query((start + end) / 2 + 1, end, i, j, 2 * node + 1, (T + X) % GCD);
  return (X + Y);
}
void update(int start, int end, int i, int node, int ne) {
  if (start > i || end < i) return;
  if (start == end) {
    ai[start] = ne;
    for (int T = 0; T < GCD; T++) {
      if (T % ai[start] == 0)
        tim[node][T] = 2;
      else
        tim[node][T] = 1;
    }
    return;
  }
  update(start, (start + end) / 2, i, 2 * node, ne);
  update((start + end) / 2 + 1, end, i, 2 * node + 1, ne);
  for (int T = 0; T < GCD; T++) {
    tim[node][T] =
        tim[2 * node][T] + tim[2 * node + 1][(T + tim[2 * node][T]) % GCD];
  }
}
int main() {
  scanf("%d", &n);
  for (int g = 1; g <= n; g++) scanf("%d", &ai[g]);
  build(1, n, 1);
  int q;
  scanf("%d", &q);
  for (int g = 0; g < q; g++) {
    char x;
    scanf("\n%c ", &x);
    if (x == 'C') {
      int a, b;
      scanf("%d %d", &a, &b);
      update(1, n, a, 1, b);
    } else {
      int a, b;
      scanf("%d %d", &a, &b);
      printf("%d\n", query(1, n, a, b - 1, 1));
    }
  }
  return 0;
}
