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 M = 100005;
const int N = 10000005;
int read() {
  int ans = 0, op = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ans *= 10;
    ans += ch - '0';
    ch = getchar();
  }
  return ans * op;
}
struct seg {
  int tim[65];
} t[M << 2];
int n, q, x, y;
char s[5];
void pushup(int p) {
  for (int i = 0; i <= 59; i++)
    t[p].tim[i] =
        t[p << 1].tim[i] + t[p << 1 | 1].tim[(i + t[p << 1].tim[i]) % 60];
}
void build(int p, int l, int r) {
  if (l == r) {
    int x = read();
    for (int i = 0; i <= 59; i++) {
      if (i % x)
        t[p].tim[i] = 1;
      else
        t[p].tim[i] = 2;
    }
    return;
  }
  int mid = (l + r) >> 1;
  build(p << 1, l, mid), build(p << 1 | 1, mid + 1, r);
  pushup(p);
}
void modify(int p, int l, int r, int pos, int val) {
  if (l == r) {
    for (int i = 0; i <= 59; i++) {
      if (i % val)
        t[p].tim[i] = 1;
      else
        t[p].tim[i] = 2;
    }
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    modify(p << 1, l, mid, pos, val);
  else
    modify(p << 1 | 1, mid + 1, r, pos, val);
  pushup(p);
}
int query(int p, int l, int r, int kl, int kr, int ti) {
  if (l == kl && r == kr) return t[p].tim[ti % 60];
  int mid = (l + r) >> 1;
  if (kr <= mid)
    return query(p << 1, l, mid, kl, kr, ti);
  else if (kl > mid)
    return query(p << 1 | 1, mid + 1, r, kl, kr, ti);
  else {
    int g = query(p << 1, l, mid, kl, mid, ti);
    return g + query(p << 1 | 1, mid + 1, r, mid + 1, kr, (ti + g) % 60);
  }
}
int main() {
  n = read(), build(1, 1, n);
  q = read();
  for (int i = 1; i <= q; i++) {
    scanf("%s", s);
    if (s[0] == 'A')
      x = read(), y = read(), printf("%d\n", query(1, 1, n, x, y - 1, 0));
    else
      x = read(), y = read(), modify(1, 1, n, x, y);
  }
  return 0;
}
