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 n, q, A[100010];
struct Segtree {
  int left[100010 * 4], right[100010 * 4];
  int key[100010 * 4][60];
  void update(int step) {
    for (int i = 0; i < (60); i++) {
      key[step][i] = key[step * 2][i];
      key[step][i] += key[step * 2 + 1][(i + key[step * 2][i]) % 60];
    }
  }
  void build(int step, int l, int r) {
    left[step] = l;
    right[step] = r;
    if (l == r) {
      for (int i = 0; i < (60); i++) key[step][i] = (i % A[l] == 0) ? 2 : 1;
      return;
    }
    int mid = left[step] + right[step] >> 1;
    build(step * 2, l, mid);
    build(step * 2 + 1, mid + 1, r);
    update(step);
  }
  void update(int step, int pos) {
    if (left[step] == right[step]) {
      for (int i = 0; i < (60); i++) key[step][i] = i % A[pos] == 0 ? 2 : 1;
      return;
    }
    int mid = left[step] + right[step] >> 1;
    update(step * 2 + (pos <= mid ? 0 : 1), pos);
    update(step);
  }
  int ask(int step, int l, int r, int t) {
    if (left[step] == l && right[step] == r) return key[step][t % 60];
    int mid = left[step] + right[step] >> 1;
    if (r <= mid)
      return ask(step * 2, l, r, t);
    else if (mid < l)
      return ask(step * 2 + 1, l, r, t);
    else {
      int ans = 0;
      ans += ask(step * 2, l, mid, t);
      return ans + ask(step * 2 + 1, mid + 1, r, t + ans);
    }
  }
} tree;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) scanf("%d", &A[i]);
  tree.build(1, 0, n - 1);
  scanf("%d", &q);
  while (q--) {
    char type[10];
    int x, y;
    scanf("%s%d%d", type, &x, &y);
    if (type[0] == 'C') {
      A[--x] = y;
      tree.update(1, x);
      ;
    } else {
      printf("%d\n", tree.ask(1, x - 1, y - 2, 0));
    }
  }
  return 0;
}
