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;
template <class T>
inline void amin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T>
inline void amax(T &a, const T &b) {
  if (a < b) a = b;
}
struct Seg {
  int T[60];
  Seg() {}
};
Seg operator+(const Seg &x, const Seg &y) {
  Seg z;
  for (int i = 0, i_len = (60); i < i_len; i++) {
    int t = x.T[i];
    z.T[i] = t + y.T[(t + i) % 60];
  }
  return z;
}
Seg single(int a) {
  Seg s;
  if (a == -1) {
    memset(s.T, 0, sizeof s.T);
  } else {
    for (int i = 0, i_len = (60); i < i_len; i++) {
      if (i % a == 0)
        s.T[i] = 2;
      else
        s.T[i] = 1;
    }
  }
  return s;
}
struct SegTree {
  int n, m;
  vector<Seg> data;
  SegTree(int n) : n(n) {
    m = 1;
    for (; m < n;) m *= 2;
    data = vector<Seg>(m * 2);
  }
  void update(int x, int a) {
    x += m;
    data[x] = single(a);
    for (x /= 2; x; x /= 2) data[x] = data[x * 2] + data[x * 2 + 1];
  }
  Seg query(int x, int y) { return query(x, y, 1, 0, m); }
  Seg query(int x, int y, int k, int l, int r) {
    if (r <= x || y <= l) return single(-1);
    if (x <= l && r <= y) return data[k];
    return query(x, y, k * 2, l, (l + r) / 2) +
           query(x, y, k * 2 + 1, (l + r) / 2, r);
  }
};
int N;
int Q;
char buf[99];
int X, Y;
int main() {
  scanf("%d", &N);
  SegTree S(N);
  for (int i = 0, i_len = (N); i < i_len; i++) {
    int a;
    scanf("%d", &a);
    S.update(i, a);
  }
  scanf("%d", &Q);
  for (int $ = 0, $_len = (Q); $ < $_len; $++) {
    scanf("%s%d%d", buf, &X, &Y);
    if (buf[0] == 'A') {
      Seg seg = S.query(X - 1, Y - 1);
      printf("%d\n", seg.T[0]);
    } else {
      S.update(X - 1, Y);
    }
  }
  return 0;
}
