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 primero = 1 << 17;
const int tope = 1 << 18;
int n;
int tiempo[tope][60];
void inserta(int pos, int val) {
  pos += primero;
  for (int i = 0; i < 60; i++) {
    if (i % val == 0)
      tiempo[pos][i] = 2;
    else
      tiempo[pos][i] = 1;
  }
  while (pos > 1) {
    pos /= 2;
    for (int i = 0; i < 60; i++)
      tiempo[pos][i] = tiempo[2 * pos][i] +
                       tiempo[2 * pos + 1][(i + tiempo[2 * pos][i]) % 60];
  }
}
int tiempori[60];
int nexttiempori[60];
int tiempoentre(int pos0, int pos1) {
  pos0 += primero;
  pos1 += primero;
  for (int i = 0; i < 60; i++) tiempori[i] = 0;
  int t = 2;
  while (pos0 + 1 < pos1) {
    if (pos0 % 2 == 0) {
      t += tiempo[pos0 + 1][t % 60];
    }
    pos0 /= 2;
    if (pos1 % 2 == 1) {
      for (int i = 0; i < 60; i++)
        nexttiempori[i] =
            tiempo[pos1 - 1][i] + tiempori[(i + tiempo[pos1 - 1][i]) % 60];
      for (int i = 0; i < 60; i++) tiempori[i] = nexttiempori[i];
    }
    pos1 /= 2;
  }
  return t + tiempori[t % 60];
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    inserta(i, a);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    char tipo;
    int x, y;
    cin >> tipo >> x >> y;
    if (tipo == 'A') {
      cout << tiempoentre(x, y) << endl;
    } else {
      inserta(x, y);
    }
  }
}
