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;
void merge(int* c, int* a, int* b) {
  int t[60];
  for (int i = 0, ThxDem = 60; i < ThxDem; ++i)
    t[i] = a[i] + b[(i + a[i]) % 60];
  for (int i = 0, ThxDem = 60; i < ThxDem; ++i) c[i] = t[i];
}
int st[400005][60];
void st_upd(int k, int s, int e, int p, int v) {
  if (s + 1 == e) {
    for (int i = 0, ThxDem = 60; i < ThxDem; ++i) st[k][i] = 1 + (i % v == 0);
  } else {
    int m = (s + e) / 2;
    if (p < m)
      st_upd(2 * k, s, m, p, v);
    else
      st_upd(2 * k + 1, m, e, p, v);
    merge(st[k], st[2 * k], st[2 * k + 1]);
  }
}
void st_query(int k, int s, int e, int a, int b, int* r) {
  if (e <= a || s >= b) return;
  if (s >= a && e <= b) {
    merge(r, r, st[k]);
    return;
  }
  int m = (s + e) / 2;
  st_query(2 * k, s, m, a, b, r);
  st_query(2 * k + 1, m, e, a, b, r);
}
int n;
int main() {
  scanf("%d", &n);
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) {
    int a;
    scanf("%d", &a);
    st_upd(1, 0, n, i, a);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    char _s[4];
    int a, b;
    scanf("%s%d%d", _s, &a, &b);
    if (_s[0] == 'A') {
      a--;
      b--;
      int r[60];
      for (int i = 0, ThxDem = 60; i < ThxDem; ++i) r[i] = 0;
      st_query(1, 0, n, a, b, r);
      printf("%d\n", r[0]);
    } else {
      a--;
      st_upd(1, 0, n, a, b);
    }
  }
  return 0;
}
