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 int in() {
  int x;
  scanf("%d", &x);
  return x;
}
const int N = 1e5 + 7, M = 60;
int n, Q;
int seg[4 * N][M];
void update(int pos, int nw, int v = 1, int s = 0, int e = n) {
  if (e == s + 1) {
    for (int i = 0; i < M; i++) seg[v][i] = 1 + !(i % nw);
    return;
  }
  int mid = (s + e) >> 1, l = v << 1, r = l | 1;
  if (pos < mid)
    update(pos, nw, l, s, mid);
  else
    update(pos, nw, r, mid, e);
  for (int i = 0; i < M; i++)
    seg[v][i] = seg[l][i] + seg[r][(seg[l][i] + i) % M];
}
int query(int l, int r, int st = 0, int v = 1, int s = 0, int e = n) {
  if (s >= r || e <= l) return 0;
  if (s >= l && e <= r) return seg[v][st];
  int mid = (s + e) / 2, i = v << 1, j = i | 1;
  int x = query(l, r, st, i, s, mid);
  return x + query(l, r, (st + x) % M, j, mid, e);
}
int32_t main() {
  n = in();
  for (int i = 0; i < n; i++) update(i, in());
  Q = in();
  while (Q--) {
    char c;
    cin >> c;
    int l = in(), r = in();
    if (c == 'C')
      update(l - 1, r);
    else
      printf("%d\n", query(l - 1, r - 1));
  }
}
