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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int n0 = 1 << 17;
int n, a[n0], q;
struct Node {
  int l, r, ans[60];
  Node() {
    memset(&ans, 0, sizeof(ans));
    l = -1, r = -1;
  }
  void init(int x) {
    l = r = x;
    for (int i = 0; i < 60; i++) ans[i] = 1 + (i % a[l] == 0);
  }
};
Node merge(Node x, Node y) {
  if (x.l == -1 && x.r == -1) return y;
  Node res;
  res.l = x.l, res.r = y.r;
  for (int i = 0; i < 60; i++)
    res.ans[i] = x.ans[i] + y.ans[(i + x.ans[i]) % 60];
  return res;
}
Node t[n0 << 1];
void build(int v, int l, int r) {
  if (l == r) {
    t[v].init(l);
    return;
  }
  int m = (l + r) / 2;
  build(v << 1, l, m), build(v << 1 | 1, m + 1, r);
  t[v] = merge(t[v << 1], t[v << 1 | 1]);
}
void upd(int v, int x, int y) {
  if (t[v].l == t[v].r) {
    a[x] = y;
    t[v].init(x);
    return;
  }
  int m = (t[v].l + t[v].r) / 2;
  if (x <= m)
    upd(v << 1, x, y);
  else
    upd(v << 1 | 1, x, y);
  t[v] = merge(t[v << 1], t[v << 1 | 1]);
}
Node res;
void get(int v, int l, int r) {
  if (t[v].l > r || t[v].r < l) return;
  if (t[v].l >= l && t[v].r <= r) {
    res = merge(res, t[v]);
    return;
  }
  get(v << 1, l, r), get(v << 1 | 1, l, r);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  for (int i = 0; i < n0; i++) a[i] = 1;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  cin >> q;
  build(1, 0, n - 1);
  while (q--) {
    int x, y;
    char c;
    cin >> c >> x >> y;
    x--;
    if (c == 'C') {
      upd(1, x, y);
    } else {
      y--;
      memset(&res.ans, 0, sizeof(res.ans)), res.l = -1, res.r = -1;
      get(1, x, y - 1);
      cout << res.ans[0] << '\n';
    }
  }
}
