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 = 101000;
const int MAXS = 60;
int tree[MAXS][MAXN << 2], a[MAXN], n;
void up(int p) {
  for (register int i = 0; i < MAXS; ++i)
    tree[i][p] =
        tree[i][p << 1] + tree[(i + tree[i][p << 1]) % MAXS][p << 1 | 1];
}
int ask(int u, int l, int r, int ll, int rr, int p) {
  if (ll <= l && r <= rr) return tree[p][u];
  int res(0), m = (l + r) >> 1;
  if (ll <= m) res = ask(u << 1, l, m, ll, rr, p);
  if (m < rr) res += ask(u << 1 | 1, m + 1, r, ll, rr, (p + res) % MAXS);
  return res;
}
void modify(int u, int l, int r, int first, int p) {
  if (l == r) {
    for (register int i = 0; i < MAXS; ++i) tree[i][u] = (i % p == 0) + 1;
    return;
  }
  int m = (l + r) >> 1;
  if (first <= m)
    modify(u << 1, l, m, first, p);
  else
    modify(u << 1 | 1, m + 1, r, first, p);
  up(u);
}
void work() {
  int T, first, second;
  char ch;
  scanf("%d", &T);
  while (T--) {
    ch = getchar();
    while (ch != 'A' && ch != 'C') ch = getchar();
    if (ch == 'A') {
      scanf("%d%d", &first, &second);
      printf("%d\n", ask(1, 1, n, first, second - 1, 0));
    } else {
      scanf("%d%d", &first, &second);
      modify(1, 1, n, first, second);
    }
  }
}
void build(int u, int l, int r) {
  if (l == r) {
    for (register int i = 0; i < MAXS; ++i) tree[i][u] = (i % a[l] == 0) + 1;
    return;
  }
  int m = (l + r) >> 1;
  build(u << 1, l, m);
  build(u << 1 | 1, m + 1, r);
  up(u);
}
void init() {
  scanf("%d", &n);
  for (register int i = (1), _end_ = (n); i <= _end_; ++i) scanf("%d", &a[i]);
  build(1, 1, n);
}
int main() {
  init();
  work();
  return 0;
}
