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;
int n, q;
vector<int> a;
vector<int> t[60];
void updateNode(int v) {
  for (int rem = 0; rem < (int)(60); rem++) {
    int tm = t[rem][((v << 1) + 1)];
    t[rem][v] = tm + t[(rem + tm) % 60][((v + 1) << 1)];
  }
}
void buildTree(int v, int l, int r) {
  if (l >= r) return;
  if (l == r - 1) {
    for (int rem = 0; rem < (int)(60); rem++) {
      if (rem % a[l] == 0)
        t[rem][v] = 2;
      else
        t[rem][v] = 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  buildTree(((v << 1) + 1), l, mid);
  buildTree(((v + 1) << 1), mid, r);
  updateNode(v);
}
void update(int v, int l, int r, int pos, int val) {
  if (l == r - 1) {
    if (l != pos) throw;
    a[pos] = val;
    for (int rem = 0; rem < (int)(60); rem++) {
      if (rem % a[l] == 0)
        t[rem][v] = 2;
      else
        t[rem][v] = 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  if (pos < mid)
    update(((v << 1) + 1), l, mid, pos, val);
  else
    update(((v + 1) << 1), mid, r, pos, val);
  updateNode(v);
}
int calc(int v, int l, int r, int x, int y, int rem) {
  if (l >= r) throw;
  if (x >= y) throw;
  if (l == x && r == y) {
    return t[rem][v];
  }
  int mid = (l + r) / 2;
  if (y <= mid) {
    return calc(((v << 1) + 1), l, mid, x, y, rem);
  } else if (x >= mid) {
    return calc(((v + 1) << 1), mid, r, x, y, rem);
  } else {
    int tm = calc(((v << 1) + 1), l, mid, x, mid, rem);
    tm += calc(((v + 1) << 1), mid, r, mid, y, (rem + tm) % 60);
    return tm;
  }
}
int main() {
  cin >> n;
  a = vector<int>(n);
  for (int i = 0; i < (int)(n); i++) cin >> a[i];
  for (int rem = 0; rem < (int)(60); rem++) {
    t[rem] = vector<int>(5 * n);
  }
  buildTree(0, 0, n);
  scanf("%d\n", &q);
  for (int it = 0; it < (int)(q); it++) {
    char comm;
    int x, y;
    scanf("%c %d %d\n", &comm, &x, &y);
    if (comm == 'A') {
      printf("%d\n", calc(0, 0, n, x - 1, y - 1, 0));
    } else if (comm == 'C') {
      x--;
      update(0, 0, n, x, y);
    } else
      throw;
  }
  return 0;
}
