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 = 100010;
struct Seg_Tree {
  int l, r;
  int d[60], res[60];
} T[MAXN << 2], ans;
int n, m, a[MAXN], l, r;
char op[10];
Seg_Tree Merge(const Seg_Tree &p, const Seg_Tree &q) {
  Seg_Tree con;
  con.l = p.l, con.r = q.r;
  for (int i = 0; i < 60; i++) {
    con.d[i] = p.d[i] + q.d[p.res[i]];
    con.res[i] = q.res[p.res[i]];
  }
  return con;
}
void MakeTree(int x, int l, int r) {
  T[x].l = l, T[x].r = r;
  if (l == r) {
    for (int i = 0; i < 60; i++)
      if (i % a[l] == 0)
        T[x].d[i] = 1, T[x].res[i] = (i + 2) % 60;
      else
        T[x].d[i] = 0, T[x].res[i] = (i + 1) % 60;
    return;
  }
  int mid = (T[x].l + T[x].r) >> 1;
  MakeTree(x << 1, l, mid);
  MakeTree(x << 1 | 1, mid + 1, r);
  T[x] = Merge(T[x << 1], T[x << 1 | 1]);
}
void Change(int x, int pos, int tar) {
  if (T[x].l == T[x].r) {
    for (int i = 0; i < 60; i++)
      if (i % tar == 0)
        T[x].d[i] = 1, T[x].res[i] = (i + 2) % 60;
      else
        T[x].d[i] = 0, T[x].res[i] = (i + 1) % 60;
    return;
  }
  int mid = (T[x].l + T[x].r) >> 1;
  if (pos <= mid)
    Change(x << 1, pos, tar);
  else
    Change(x << 1 | 1, pos, tar);
  T[x] = Merge(T[x << 1], T[x << 1 | 1]);
}
Seg_Tree Query(int x, int l, int r) {
  if (T[x].l == l && T[x].r == r) return T[x];
  int mid = (T[x].l + T[x].r) >> 1;
  if (r <= mid)
    return Query(x << 1, l, r);
  else if (l > mid)
    return Query(x << 1 | 1, l, r);
  else
    return Merge(Query(x << 1, l, mid), Query(x << 1 | 1, mid + 1, r));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  MakeTree(1, 1, n);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s%d%d", op, &l, &r);
    if (op[0] == 'C')
      Change(1, l, r);
    else {
      ans = Query(1, l, r - 1);
      printf("%d\n", r - l + ans.d[0]);
    }
  }
  return 0;
}
