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>
#pragma comment(linker, "/stack:16777216")
using namespace std;
const int INF = 1000000000;
const int MAX = 100007;
const int MAX2 = 10007;
const int BASE = 1000000000;
const int MOD = 1000000007;
int t[61][4 * MAX];
int a[MAX];
void build(int v, int l, int r) {
  if (l == r) {
    for (int i = (0); i < (60); ++i) {
      if (i % a[l] == 0)
        t[i][v] = 2;
      else
        t[i][v] = 1;
    }
    return;
  }
  int m = (l + r) / 2;
  build(2 * v + 1, l, m);
  build(2 * v + 2, m + 1, r);
  for (int i = (0); i < (60); ++i) {
    t[i][v] = t[i][2 * v + 1] + t[(i + t[i][2 * v + 1]) % 60][2 * v + 2];
  }
}
int res;
void get(int v, int l, int r, int L, int R) {
  if (l == L && r == R) {
    res += t[res % 60][v];
    return;
  }
  int m = (l + r) / 2;
  if (L <= m) {
    get(2 * v + 1, l, m, L, min(R, m));
  }
  if (R >= m + 1) {
    get(2 * v + 2, m + 1, r, max(L, m + 1), R);
  }
}
void upd(int v, int l, int r, int ind, int val) {
  if (l == r) {
    a[l] = val;
    for (int i = (0); i < (60); ++i) {
      if (i % a[l] == 0)
        t[i][v] = 2;
      else
        t[i][v] = 1;
    }
    return;
  }
  int m = (l + r) / 2;
  if (ind <= m) {
    upd(2 * v + 1, l, m, ind, val);
  } else {
    upd(2 * v + 2, m + 1, r, ind, val);
  }
  for (int i = (0); i < (60); ++i) {
    t[i][v] = t[i][2 * v + 1] + t[(i + t[i][2 * v + 1]) % 60][2 * v + 2];
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = (0); i < (n); ++i) {
    scanf("%d", &a[i]);
  }
  int q;
  cin >> q;
  build(0, 0, n - 1);
  for (int i = (0); i < (q); ++i) {
    char ch;
    scanf("%c", &ch);
    int x, y;
    scanf("%c %d %d", &ch, &x, &y);
    if (ch == 'A') {
      --x;
      y -= 2;
      res = 0;
      get(0, 0, n - 1, x, y);
      printf("%d\n", res);
    }
    if (ch == 'C') {
      --x;
      upd(0, 0, n - 1, x, y);
    }
  }
  return 0;
}
