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 MN = 100111;
int a[MN];
struct Node {
  int go[60];
} it[MN * 4];
Node operator+(const Node& a, const Node& b) {
  Node res;
  for (int x = 0, _a = (60); x < _a; x++) {
    res.go[x] = a.go[x] + b.go[(x + a.go[x]) % 60];
  }
  return res;
}
Node init(int a) {
  Node res;
  for (int x = 0, _a = (60); x < _a; x++) {
    res.go[x] = (x % a) ? 1 : 2;
  }
  return res;
}
void build(int i, int l, int r) {
  if (l == r) {
    it[i] = init(a[l]);
    return;
  }
  int mid = (l + r) >> 1;
  build(((i) << 1), l, mid);
  build((((i) << 1) + 1), mid + 1, r);
  it[i] = it[((i) << 1)] + it[(((i) << 1) + 1)];
}
void update(int i, int l, int r, int u, int val) {
  if (u < l || r < u) return;
  if (l == r) {
    it[i] = init(val);
    return;
  }
  int mid = (l + r) >> 1;
  update(((i) << 1), l, mid, u, val);
  update((((i) << 1) + 1), mid + 1, r, u, val);
  it[i] = it[((i) << 1)] + it[(((i) << 1) + 1)];
}
Node get(int i, int l, int r, int u, int v) {
  if (u <= l && r <= v) return it[i];
  int mid = (l + r) >> 1;
  if (v <= mid) return get(((i) << 1), l, mid, u, v);
  if (mid + 1 <= u) return get((((i) << 1) + 1), mid + 1, r, u, v);
  Node left = get(((i) << 1), l, mid, u, v);
  Node right = get((((i) << 1) + 1), mid + 1, r, u, v);
  return left + right;
}
int main() {
  ios ::sync_with_stdio(false);
  cout << (fixed) << setprecision(9);
  int n;
  while (cin >> n) {
    for (int i = (1), _b = (n); i <= _b; i++) cin >> a[i];
    build(1, 1, n);
    int q;
    cin >> q;
    while (q--) {
      char typ;
      cin >> typ;
      if (typ == 'C') {
        int u, val;
        cin >> u >> val;
        update(1, 1, n, u, val);
      } else {
        int u, v;
        cin >> u >> v;
        --v;
        Node res = get(1, 1, n, u, v);
        cout << res.go[0] << "\n";
      }
    }
  }
  return 0;
}
