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 = 100005;
int a[N];
struct Seg {
  int l, r, mid;
  int t[60];
} seg[3 * N];
void up(int n) {
  for (int i = 0; i < 60; ++i) {
    int tmp = i + seg[2 * n].t[i];
    seg[n].t[i] = tmp - i + seg[2 * n + 1].t[tmp % 60];
  }
}
void build(int l, int r, int n) {
  seg[n].l = l;
  seg[n].r = r;
  seg[n].mid = (l + r) >> 1;
  if (l == r) {
    for (int i = 0; i < 60; ++i)
      if (i % a[l] == 0)
        seg[n].t[i] = 2;
      else
        seg[n].t[i] = 1;
    return;
  }
  build(l, seg[n].mid, 2 * n);
  build(seg[n].mid + 1, r, 2 * n + 1);
  up(n);
}
void update(int u, int n, int x) {
  if (seg[n].l == seg[n].r) {
    a[u] = x;
    for (int i = 0; i < 60; ++i)
      if (i % a[u] == 0)
        seg[n].t[i] = 2;
      else
        seg[n].t[i] = 1;
    return;
  }
  if (u <= seg[n].mid)
    update(u, 2 * n, x);
  else
    update(u, 2 * n + 1, x);
  up(n);
}
int query(int l, int r, int n, int t) {
  if (l <= seg[n].l && seg[n].r <= r) {
    return t + seg[n].t[t % 60];
  }
  if (seg[n].mid >= l) t = query(l, r, 2 * n, t);
  if (seg[n].mid < r) t = query(l, r, 2 * n + 1, t);
  return t;
}
void debug(int n) {
  cout << "---" << endl;
  for (int i = 1; i <= n; ++i) cout << query(i, i, 1, 0) << ' ';
  cout << endl;
  cout << "---" << endl;
}
void debug1(int n) {
  cout << seg[n].l << ' ' << seg[n].r << ' ' << seg[n].t[0] << endl;
  if (seg[n].l == seg[n].r) return;
  debug1(2 * n);
  debug1(2 * n + 1);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  build(1, n, 1);
  int m;
  scanf("%d", &m);
  char op[2];
  int u, v;
  for (int i = 0; i < m; ++i) {
    scanf("%s%d%d", op, &u, &v);
    if (op[0] == 'A') {
      printf("%d\n", query(u, v - 1, 1, 0));
    } else {
      update(u, 1, v);
    }
  }
  return 0;
}
