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() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const long long infll = 1e18 + 10;
const int inf = 1e9;
int a[maxn], n;
int stree[60][4 * maxn];
inline void build(int p = 0, int l = 0, int r = n - 1) {
  if (l == r) {
    for (int pt = 0; pt < 60; pt++) {
      stree[pt][p] = (pt % a[l] == 0 ? 2 : 1);
    }
    return;
  }
  int mid = (l + r) >> 1;
  build(2 * p + 1, l, mid), build(2 * p + 2, mid + 1, r);
  for (int i = 0; i < 60; i++)
    stree[i][p] = stree[((stree[i][2 * p + 1]) + i) % 60][2 * p + 2] +
                  stree[i][2 * p + 1];
}
inline pair<int, int> get(int ql = 0, int qr = 0, pair<int, int> ret = {0, 0},
                          int p = 0, int l = 0, int r = n - 1) {
  if (l > qr || r < ql) return ret;
  if (l >= ql && r <= qr)
    return {(stree[ret.first][p] + ret.first) % 60,
            ret.second + stree[ret.first][p]};
  int mid = (l + r) >> 1;
  pair<int, int> start = get(ql, qr, ret, 2 * p + 1, l, mid);
  return get(ql, qr, start, 2 * p + 2, mid + 1, r);
}
inline void update(int qx, int p = 0, int l = 0, int r = n - 1) {
  if (l > qx || r < qx) return;
  if (l == r) {
    for (int pt = 0; pt < 60; pt++) {
      stree[pt][p] = (pt % a[l] == 0 ? 2 : 1);
    }
    return;
  }
  int mid = (l + r) >> 1;
  update(qx, 2 * p + 1, l, mid), update(qx, 2 * p + 2, mid + 1, r);
  for (int i = 0; i < 60; i++)
    stree[i][p] = stree[((stree[i][2 * p + 1]) + i) % 60][2 * p + 2] +
                  stree[i][2 * p + 1];
}
int32_t main() {
  n = in();
  for (int i = 0; i < n; i++) a[i] = in();
  build();
  int q = in();
  while (q--) {
    char ch = 0;
    do {
      ch = getchar();
    } while (!isalpha(ch));
    if (ch == 'A') {
      int x = in() - 1, y = in() - 2;
      cout << get(x, y).second << "\n";
    } else {
      int x = in() - 1, val = in();
      a[x] = val;
      update(x);
    }
  }
}
