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 = 300000 + 20;
int n, m, a[MAXN];
struct seg {
  int l, r;
  int t[60];
} tr[MAXN * 3];
seg merge(seg a, seg b) {
  seg tmp;
  tmp.l = a.l;
  tmp.r = b.r;
  for (int i = 0; i < 60; i++) tmp.t[i] = a.t[i] + b.t[(i + a.t[i]) % 60];
  return tmp;
}
void upp(int p) { tr[p] = merge(tr[(p << 1)], tr[((p << 1) | 1)]); }
void build(int p, int l, int r) {
  tr[p].l = l;
  tr[p].r = r;
  if ((tr[p].l == tr[p].r)) {
    int val;
    scanf("%d", &val);
    for (int i = 0; i < 60; i++) tr[p].t[i] = ((i % val) == 0) ? 2 : 1;
    return;
  }
  int mid = (l + r) >> 1;
  build((p << 1), l, mid);
  build(((p << 1) | 1), mid + 1, r);
  upp(p);
}
void ins(int p, int x, int val) {
  int l = tr[p].l, r = tr[p].r;
  if ((tr[p].l == tr[p].r)) {
    for (int i = 0; i < 60; i++) tr[p].t[i] = ((i % val) == 0) ? 2 : 1;
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    ins((p << 1), x, val);
  else
    ins(((p << 1) | 1), x, val);
  upp(p);
}
seg qry(int p, int x, int y) {
  int l = tr[p].l, r = tr[p].r;
  if (((l == x) && (y == r))) return tr[p];
  int mid = (l + r) >> 1;
  if (y <= mid)
    return qry((p << 1), x, y);
  else if (mid < x)
    return qry(((p << 1) | 1), x, y);
  else
    return merge(qry((p << 1), x, mid), qry(((p << 1) | 1), mid + 1, y));
}
int main() {
  scanf("%d", &n);
  build(1, 1, n);
  scanf("%d", &m);
  char s[5];
  int x, y;
  while (m--) {
    scanf("%s", s);
    scanf("%d%d", &x, &y);
    if (s[0] == 'A') {
      printf("%d\n", qry(1, x, y - 1).t[0]);
    } else {
      ins(1, x, y);
    }
  }
  return 0;
}
