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;
struct T {
  int v[60];
};
T m[2 * (1 << 17)];
void upd(T &h, T &l, T &r) {
  for (int i = 0; i < 60; h.v[i] = l.v[i] + r.v[(i + l.v[i]) % 60], i++)
    ;
}
void make(T &h, int v) {
  for (int i = 0; i < 60; h.v[i] = 1 + (i % v == 0), i++)
    ;
}
void upd(int i) {
  for (i = (i + (1 << 17)) >> 1; i; upd(m[i], m[2 * i], m[2 * i + 1]), i >>= 1)
    ;
}
int gettime(int i, int j) {
  int t, a[20], c;
  t = 0;
  c = 0;
  for (i += (1 << 17), j += (1 << 17); i <= j;
       i = (i + 1) >> 1, j = (j - 1) >> 1) {
    if (i & 1) t += m[i].v[t % 60];
    if (~j & 1) a[c++] = j;
  }
  for (; c--; t += m[a[c]].v[t % 60])
    ;
  return t;
}
int main() {
  int n, i, j, q;
  scanf("%d", &n);
  for (i = 0; i < n; scanf("%d", &j), make(m[(1 << 17) + i], j), i++)
    ;
  for (i = (1 << 17) - 1; i > 0; upd(m[i], m[2 * i], m[2 * i + 1]), i--)
    ;
  scanf("%d", &q);
  for (; q--;) {
    char e[3];
    scanf("%s%d%d", e, &i, &j);
    if (e[0] == 'A')
      printf("%d\n", gettime(i - 1, j - 2));
    else {
      make(m[(1 << 17) + i - 1], j);
      upd(i - 1);
    }
  }
  return 0;
}
