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 MAXN = 1e5 + 5, V = 60, MAXS = 4 * MAXN;
int n, a[MAXN], q, x, y;
char qt;
int tree[V][MAXS];
void upd(int v, int b, int e, int val) {
  int m = b + e >> 1;
  for (int i = 0; i < V; ++i) {
    if (b == e)
      tree[i][v] = i % val == 0 ? 1 : 0;
    else {
      int tl = tree[i][v + v];
      int rr = ((m - b + 1) + tl);
      int tr = tree[(i + rr) % V][v + v + 1];
      tree[i][v] = tl + tr;
    }
  }
}
void build(int v, int b, int e) {
  int m = b + e >> 1;
  if (b != e) {
    build(v + v, b, m);
    build(v + v + 1, m + 1, e);
  }
  upd(v, b, e, a[b]);
}
int query(int v, int b, int e, int l, int r, int rem) {
  if (l <= b && e <= r) return tree[rem][v] + (e - b + 1);
  if (e < l || b > r) return 0;
  int m = b + e >> 1;
  int tl = query(v + v, b, m, l, r, rem);
  int tr = query(v + v + 1, m + 1, e, l, r, (rem + tl) % V);
  return tl + tr;
}
void update(int v, int b, int e, int pos, int val) {
  int m = b + e >> 1;
  if (b != e) {
    if (pos <= m)
      update(v + v, b, m, pos, val);
    else
      update(v + v + 1, m + 1, e, pos, val);
  }
  upd(v, b, e, val);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  build(1, 1, n);
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> qt >> x >> y;
    if (qt == 'A') {
      cout << query(1, 1, n, x, y - 1, 0) << '\n';
    } else {
      update(1, 1, n, x, y);
    }
  }
  return 0;
}
