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;
inline void read(int &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
inline void read(long long &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
inline void write(int x) {
  static const int maxlen = 100;
  static char s[maxlen];
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (!x) {
    putchar('0');
    return;
  }
  int len = 0;
  for (; x; x /= 10) s[len++] = x % 10 + '0';
  for (int i = len - 1; i >= 0; --i) putchar(s[i]);
}
const int MAXN = 100010;
const int MAXP = 60;
int n, m;
int a[MAXN];
char s[100];
int nex[MAXN];
int get_nex(int x) {
  int now = 0;
  for (int i = x; i <= n; i++) {
    now += (now % a[i]) ? 1 : 2;
    if (now == 60) return i + 1;
  }
  return n + 1;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= n; i++) nex[i] = get_nex(i);
  read(m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", s);
    int x, y;
    read(x);
    read(y);
    if (s[0] == 'A') {
      int ans = 0;
      while (nex[x] < y) x = nex[x], ans += 60;
      while (x < y) {
        ans += (ans % a[x]) ? 1 : 2;
        x++;
      }
      printf("%d\n", ans);
    } else {
      a[x] = y;
      for (int i = x; nex[i] >= x; i--) nex[i] = get_nex(i);
    }
  }
  return 0;
}
