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;
template <class T, class U>
void ckmin(T &a, U b) {
  if (a > b) a = b;
}
template <class T, class U>
void ckmax(T &a, U b) {
  if (a < b) a = b;
}
int N, Q;
int arr[100013];
array<int, 60> seg[3 * 100013];
array<int, 60> base[7];
int ans;
array<int, 60> comb(array<int, 60> l, array<int, 60> r) {
  array<int, 60> res;
  for (auto i = (0); i < (60); i++) {
    res[i] = l[i] + r[(i + l[i]) % 60];
  }
  return res;
}
void build(int w, int L, int R) {
  if (L == R) {
    seg[w] = base[arr[L]];
    return;
  }
  int mid = (L + R) >> 1;
  build(w << 1, L, mid);
  build(w << 1 | 1, mid + 1, R);
  seg[w] = comb(seg[w << 1], seg[w << 1 | 1]);
}
void update(int w, int L, int R, int a, int v) {
  if (L == R) {
    seg[w] = base[v];
    return;
  }
  int mid = (L + R) >> 1;
  if (a <= mid)
    update(w << 1, L, mid, a, v);
  else
    update(w << 1 | 1, mid + 1, R, a, v);
  seg[w] = comb(seg[w << 1], seg[w << 1 | 1]);
}
array<int, 60> query(int w, int L, int R, int a, int b) {
  if (a <= L && R <= b) {
    return seg[w];
  }
  int mid = (L + R) >> 1;
  if (b <= mid) return query(w << 1, L, mid, a, b);
  if (mid < a) return query(w << 1 | 1, mid + 1, R, a, b);
  return comb(query(w << 1, L, mid, a, b), query(w << 1 | 1, mid + 1, R, a, b));
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (auto i = (0); i < (N); i++) cin >> arr[i];
  for (auto i = (2); i < (7); i++) {
    for (auto j = (0); j < (60); j++) {
      base[i][j] = (j % i ? 1 : 2);
    }
  }
  build(1, 0, N - 1);
  cin >> Q;
  while (Q--) {
    char qid;
    int l, r;
    cin >> qid >> l >> r;
    if (qid == 'C') {
      l--;
      arr[l] = r;
      update(1, 0, N - 1, l, r);
    }
    if (qid == 'A') {
      l--;
      r--;
      array<int, 60> res = query(1, 0, N - 1, l, r - 1);
      cout << res[0] << '\n';
    }
  }
  return 0;
}
