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;
#pragma warning(disable : 4996)
int N, Q, a[130000], t[60][2 * 130000];
int merge(int time, int n) {
  int ret = t[time][n << 1];
  return ret + t[(time + ret) % 60][n << 1 | 1];
}
int query(int n, int time, int l, int r, int size) {
  if (l >= r) return 0;
  if (!l && r == size) return t[time][n];
  size >>= 1;
  int ret = query(n << 1, time, l, min(size, r), size);
  return ret +
         query(n << 1 | 1, (time + ret) % 60, max(0, l - size), r - size, size);
}
int main() {
  while (scanf("%d", &N) != EOF) {
    for (int i = int(0); i < int(N); i++) scanf("%d", a + i);
    int ts = 1 << (int)ceil(log2(N));
    for (int i = int(0); i < int(N); i++) {
      for (int j = int(0); j < int(60); j++) t[j][ts + i] = 1 + !(j % a[i]);
    }
    for (int i = ts - 1; i; i--) {
      for (int j = int(0); j < int(60); j++) {
        t[j][i] = merge(j, i);
      }
    }
    scanf("%d", &Q);
    char s[2];
    int x, y;
    for (int i = int(0); i < int(Q); i++) {
      scanf("%s %d %d", s, &x, &y);
      if (s[0] == 'C') {
        x += ts - 1;
        for (int j = int(0); j < int(60); j++) {
          t[j][x] = 1 + !(j % y);
        }
        while (x >>= 1) {
          for (int j = int(0); j < int(60); j++) {
            t[j][x] = merge(j, x);
          }
        }
      } else {
        printf("%d\n", query(1, 0, x - 1, y - 1, ts));
      }
    }
  }
  return 0;
}
