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 = 1e5 + 9, P = 60;
template <typename V>
struct T {
  V c[N * 4];
  void build(int p, int L, int R, V *a) {
    if (L == R) {
      c[p] = a[L];
      return;
    }
    build((p << 1), L, ((L + R) >> 1), a);
    build(((p << 1) | 1), ((L + R) >> 1) + 1, R, a);
    c[p] = c[(p << 1)] + c[((p << 1) | 1)];
  }
  V query(int p, int L, int R, int l, int r) {
    if (l == L && r == R) {
      return c[p];
    }
    if (r <= ((L + R) >> 1)) {
      return query((p << 1), L, ((L + R) >> 1), l, r);
    } else if (l > ((L + R) >> 1)) {
      return query(((p << 1) | 1), ((L + R) >> 1) + 1, R, l, r);
    } else {
      return query((p << 1), L, ((L + R) >> 1), l, ((L + R) >> 1)) +
             query(((p << 1) | 1), ((L + R) >> 1) + 1, R, ((L + R) >> 1) + 1,
                   r);
    }
  }
  void modify(int p, int L, int R, int id, const V &x) {
    if (L == R) {
      c[p] = x;
      return;
    }
    if (id <= ((L + R) >> 1))
      modify((p << 1), L, ((L + R) >> 1), id, x);
    else
      modify(((p << 1) | 1), ((L + R) >> 1) + 1, R, id, x);
    c[p] = c[(p << 1)] + c[((p << 1) | 1)];
  }
};
struct D {
  int a[P];
  D operator+(const D &o) const {
    D r;
    for (int i = 0; i < P; ++i) r.a[i] = a[i] + o.a[(i + a[i]) % P];
    return r;
  }
  void set(int x) {
    for (int i = 0; i < P; ++i) {
      a[i] = 1 + (i % x == 0);
    }
  }
};
int n;
D a[N];
T<D> t;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    a[i].set(x);
  }
  t.build(1, 1, n, a);
  int q;
  scanf("%d", &q);
  while (q--) {
    char ch[4];
    int x, y;
    scanf("%s%d%d", ch, &x, &y);
    if (ch[0] == 'A') {
      D ans;
      ans = t.query(1, 1, n, x, y - 1);
      printf("%d\n", ans.a[0]);
    } else {
      D tp;
      tp.set(y);
      t.modify(1, 1, n, x, tp);
    }
  }
  return 0;
}
