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 N = 100100, V = 60, L = 0, R = 1;
struct node {
  int l, r, to[V], stops[V], s[2];
} t[N + N] = {};
int n, a[N] = {}, tot = 0, root = 0;
inline void update(int k) {
  int k1 = t[k].s[L], k2 = t[k].s[R];
  for (int i = 0; i < V; ++i) {
    int j = t[k1].to[i];
    t[k].to[i] = t[k2].to[j];
    t[k].stops[i] = t[k1].stops[i] + t[k2].stops[j];
  }
}
inline void calc(int k) {
  int l = t[k].l;
  for (int i = 0; i < V; ++i)
    if (i % a[l] == 0)
      t[k].to[i] = (i + 2) % V, t[k].stops[i] = 1;
    else
      t[k].to[i] = (i + 1) % V, t[k].stops[i] = 0;
}
int make_tree(int l, int r) {
  int k = ++tot;
  t[k].l = l;
  t[k].r = r;
  if (l != r) {
    int mid = (l + r) >> 1;
    t[k].s[L] = make_tree(l, mid);
    t[k].s[R] = make_tree(mid + 1, r);
    update(k);
  } else
    calc(k);
  return k;
}
void change(int pos, int k) {
  if (t[k].l == t[k].r)
    calc(k);
  else {
    t[t[k].s[L]].r >= pos ? change(pos, t[k].s[L]) : change(pos, t[k].s[R]);
    update(k);
  }
}
int get_stops(int l, int r, int k, int &now) {
  if (t[k].l >= l && t[k].r <= r) {
    int ans = t[k].stops[now];
    now = t[k].to[now];
    return ans;
  }
  int s = 0;
  if (t[t[k].s[L]].r >= l) s += get_stops(l, r, t[k].s[L], now);
  if (t[t[k].s[R]].l <= r) s += get_stops(l, r, t[k].s[R], now);
  return s;
}
void init() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  root = make_tree(1, n);
}
void work() {
  int q = 0, x, y;
  char c;
  cin >> q;
  for (int i = 1; i <= q; ++i) {
    cin >> c >> x >> y;
    if (c == 'C') {
      a[x] = y;
      change(x, root);
    } else {
      int tmp = 0;
      printf("%d\n", y - x + get_stops(x, y - 1, root, tmp));
    }
  }
}
int main() {
  init();
  work();
  return 0;
}
