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 <typename T>
inline bool smin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline bool smax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const int N = 1e5 + 10, maxn = 60;
int n, ar[N], q;
struct node {
  int t[maxn + 10];
  node() { memset(t, 0, sizeof t); }
  inline void cls() { memset(t, 0, sizeof t); }
} seg[N << 2];
inline node mrg(node a, node b) {
  node ret;
  ret.cls();
  for (int i = 0; i < maxn; ++i) {
    int current = a.t[i];
    current += b.t[(i + current) % maxn];
    ret.t[i] = current;
  }
  return ret;
}
void update(int pos, int val, int id = 1, int l = 0, int r = n) {
  if (r - l < 2) {
    ar[pos] = val;
    for (int i = 0; i < maxn; ++i) {
      seg[id].t[i] = ((i % val) == 0) + 1;
    }
    return;
  }
  int mid = (r + l) >> 1, lc = id << 1, rc = lc | 1;
  if (pos < mid)
    update(pos, val, lc, l, mid);
  else
    update(pos, val, rc, mid, r);
  seg[id] = mrg(seg[lc], seg[rc]);
}
int get(int x, int y, int cur = 0, int id = 1, int l = 0, int r = n) {
  if (r <= x || l >= y) return 0;
  if (l >= x && r <= y) return seg[id].t[cur];
  int mid = (r + l) >> 1, lc = id << 1, rc = lc | 1;
  int now = get(x, y, cur, lc, l, mid);
  now += get(x, y, (cur + now) % maxn, rc, mid, r);
  return now;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> ar[i];
  for (int i = 0; i < n; ++i) update(i, ar[i]);
  cin >> q;
  while (q--) {
    char c;
    cin >> c;
    if (c == 'C') {
      int pos, val;
      cin >> pos >> val;
      update(--pos, val);
    } else {
      int u, v;
      cin >> u >> v;
      cout << get(--u, --v) << endl;
    }
  }
}
