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 mod = 60;
const int maxN = (1e5) + 5;
int A[4 * maxN + 5][mod + 1];
int n, q, v[maxN], x, y;
char c;
void build(int nod, int st, int dr) {
  if (st == dr) {
    for (int i = 0; i < mod; i++) {
      int t = 0;
      if (!(i % v[st])) t++;
      t++;
      A[nod][i] = t;
    }
    return;
  }
  int mid = (st + dr) >> 1;
  build(2 * nod, st, mid);
  build(2 * nod + 1, mid + 1, dr);
  for (int i = 0; i < mod; i++) {
    int newrem = (i + A[2 * nod][i]) % mod;
    A[nod][i] = (A[2 * nod][i] + A[2 * nod + 1][newrem]);
  }
}
int query(int nod, int st, int dr, int a, int b, int rest) {
  if (a <= st && dr <= b) {
    return A[nod][rest];
  }
  int mid = (st + dr) >> 1;
  int sol = 0;
  if (a <= mid && b <= mid)
    sol = query(2 * nod, st, mid, a, b, rest);
  else if (a > mid && b > mid)
    sol = query(2 * nod + 1, mid + 1, dr, a, b, rest);
  else {
    int newrem = (rest + query(2 * nod, st, mid, a, b, rest)) % mod;
    sol = (query(2 * nod, st, mid, a, b, rest) +
           query(2 * nod + 1, mid + 1, dr, a, b, newrem));
  }
  return sol;
}
void update(int nod, int st, int dr, int pos, int val) {
  if (st == dr) {
    v[st] = val;
    for (int i = 0; i < mod; i++) {
      int t = 0;
      if (!(i % v[st])) t++;
      t++;
      A[nod][i] = t;
    }
    return;
  }
  int mid = (st + dr) >> 1;
  if (pos <= mid)
    update(2 * nod, st, mid, pos, val);
  else
    update(2 * nod + 1, mid + 1, dr, pos, val);
  for (int i = 0; i < mod; i++) {
    int newrem = (i + A[2 * nod][i]) % mod;
    A[nod][i] = (A[2 * nod][i] + A[2 * nod + 1][newrem]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
  build(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    scanf("\n");
    scanf("%c", &c);
    if (c == 'A') {
      scanf("%d%d", &x, &y);
      printf("%d\n", query(1, 1, n, x, y - 1, 0));
    } else {
      int pos, val;
      scanf("%d%d", &pos, &val);
      update(1, 1, n, pos, val);
    }
  }
  return 0;
}
