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;
const int maxn = 100005;
struct Node {
  int t[60];
} v[maxn << 2];
int n, Q, A[maxn];
void build(int l, int r, int x = 1) {
  if (l == r) return;
  int m = (l + r) >> 1;
  build(l, m, x << 1), build(m + 1, r, x << 1 | 1);
  for (int i = 0; i < (int)(60); ++i) {
    int tmp;
    v[x].t[i] = v[x << 1].t[i] + 1 + ((v[x << 1].t[i] + i) % A[m] == 0),
    v[x].t[i] += tmp = v[x << 1 | 1].t[(v[x].t[i] + i) % 60];
  }
}
int q_res, pre, prel, prer;
void query(int l, int r, int L, int R, int x = 1) {
  if (L <= l && r <= R) {
    if (pre == -1)
      q_res = v[x].t[0];
    else
      q_res += 1 + (q_res % A[pre] == 0), q_res += v[x].t[q_res % 60];
    pre = r, prel = l, prer = r;
    return;
  }
  int m = (l + r) >> 1;
  if (L <= m) query(l, m, L, R, x << 1);
  if (R > m) query(m + 1, r, L, R, x << 1 | 1);
}
void modify(int l, int r, int pos, int x = 1) {
  if (l == r) return;
  int m = (l + r) >> 1;
  if (pos <= m)
    modify(l, m, pos, x << 1);
  else
    modify(m + 1, r, pos, x << 1 | 1);
  for (int i = 0; i < (int)(60); ++i)
    v[x].t[i] = v[x << 1].t[i] + 1 + ((v[x << 1].t[i] + i) % A[m] == 0),
    v[x].t[i] += v[x << 1 | 1].t[(v[x].t[i] + i) % 60];
}
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  for (int i = (int)(1); i <= (int)(n); ++i) scanf("%d", A + i);
  build(1, n + 1);
  scanf("%d", &Q);
  for (int q = 0; q < (int)(Q); ++q) {
    int x, y;
    char op[3];
    scanf("%s%d%d", op, &x, &y);
    if (op[0] == 'A') {
      pre = -1, q_res = 0;
      query(1, n + 1, x, y, 1);
      printf("%d\n", q_res);
    } else {
      A[x] = y;
      modify(1, n + 1, x, 1);
    }
  }
  return 0;
}
