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 value[100005];
int AI[60][4 * 100005];
void build(int node, int left, int right) {
  int m = (left + right) / 2;
  if (left == right) {
    for (int i = 0; i < 60; i++) {
      if (i % value[left] == 0)
        AI[i][node] = 2;
      else
        AI[i][node] = 1;
    }
    return;
  }
  build(node * 2, left, m);
  build(node * 2 + 1, m + 1, right);
  for (int i = 0; i < 60; i++) {
    int s = AI[i][node * 2];
    s += AI[(i + s) % 60][node * 2 + 1];
    AI[i][node] = s;
  }
}
void update(int node, int left, int right, int position) {
  int m = (left + right) / 2;
  if (left == right) {
    for (int i = 0; i < 60; i++) {
      if (i % value[left] == 0)
        AI[i][node] = 2;
      else
        AI[i][node] = 1;
    }
    return;
  }
  if (m >= position)
    update(node * 2, left, m, position);
  else
    update(node * 2 + 1, m + 1, right, position);
  for (int i = 0; i < 60; i++) {
    int s = AI[i][node * 2];
    s += AI[(i + s) % 60][node * 2 + 1];
    AI[i][node] = s;
  }
}
int query(int node, int left, int right, int a, int b, int start) {
  int m = (left + right) / 2, total = 0;
  if (a <= left && b >= right) return AI[start][node];
  if (m >= a) total = query(node * 2, left, m, a, b, start);
  if (m < b)
    total += query(node * 2 + 1, m + 1, right, a, b, (start + total) % 60);
  return total;
}
int main() {
  int N;
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> value[i];
  build(1, 1, N);
  int t;
  cin >> t;
  while (t--) {
    char ch;
    int x, y;
    cin >> ch >> x >> y;
    if (ch == 'C') {
      value[x] = y;
      update(1, 1, N, x);
    } else
      cout << query(1, 1, N, x, y - 1, 0) << "\n";
  }
}
