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 INF = 1e9;
const long long int INFLL = 1e18;
const double EPS = 1e-8;
template <class T>
T &chmin(T &a, const T &b) {
  return a = min(a, b);
}
template <class T>
T &chmax(T &a, const T &b) {
  return a = max(a, b);
}
const int MAX_N = 100010;
int size;
int node[MAX_N * 4][60];
void init(int n) {
  size = 1;
  while (size < n) size *= 2;
  for (int i = (0); i < (int)(size * 2 - 1); i++)
    for (int j = (0); j < (int)(60); j++) node[i][j] = 0;
}
void update(int k, int x) {
  k += size - 1;
  for (int i = (0); i < (int)(60); i++) {
    if (i % x == 0)
      node[k][i] = 2;
    else
      node[k][i] = 1;
  }
  while (k > 0) {
    k = (k - 1) / 2;
    for (int i = (0); i < (int)(60); i++)
      node[k][i] =
          node[k * 2 + 1][i] + node[k * 2 + 2][(i + node[k * 2 + 1][i]) % 60];
  }
}
int query(int a, int b, int t, int k, int l, int r) {
  if (a <= l && r <= b) {
    return node[k][t % 60];
  } else if (b <= l || r <= a) {
    return 0;
  } else {
    int chl = query(a, b, t, k * 2 + 1, l, (l + r) / 2);
    int chr = query(a, b, t + chl, k * 2 + 2, (l + r) / 2, r);
    return chl + chr;
  }
}
int query(int a, int b, int t) { return query(a, b, t, 0, 0, size); }
int n;
int a[MAX_N];
int q;
char c;
int x, y;
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (int)(n); i++) scanf("%d", &a[i]);
  init(n);
  for (int i = (0); i < (int)(n); i++) update(i, a[i]);
  scanf("%d\n", &q);
  for (int loop_cnt = (0); loop_cnt < (int)(q); loop_cnt++) {
    scanf("%c %d %d\n", &c, &x, &y);
    if (c == 'C') {
      x--;
      update(x, y);
    } else if (c == 'A') {
      x--;
      y--;
      printf("%d\n", query(x, y, 0));
    }
  }
  return 0;
}
