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 M = 400010, mod = 1000000007;
int T[M][62];
int A[M];
void merhe(int i, int l, int r) {
  for (int t = 0; t <= 60; t++) {
    T[i][t] = T[l][t] + T[r][(t + T[l][t]) % 60];
  }
}
void build(int i, int l, int r) {
  if (l == r) {
    for (int t = 0; t <= 60; t++) {
      if (t % A[l] == 0)
        T[i][t] = 2;
      else
        T[i][t] = 1;
    }
  } else {
    int m = (l + r) / 2;
    build((2 * i), l, m);
    build((2 * i + 1), m + 1, r);
    merhe(i, (2 * i), (2 * i + 1));
  }
}
void update(int i, int l, int r, int first, int val) {
  if (l == r && r == first) {
    A[l] = val;
    for (int t = 0; t <= 60; t++) {
      if (t % A[l] == 0)
        T[i][t] = 2;
      else
        T[i][t] = 1;
    }
  } else if (l <= first && first <= r) {
    int m = (l + r) / 2;
    update((2 * i), l, m, first, val);
    update((2 * i + 1), m + 1, r, first, val);
    merhe(i, (2 * i), (2 * i + 1));
  }
}
int TIME;
void query(int i, int l, int r, int st, int ed) {
  if (l >= st && r <= ed) {
    TIME = TIME + T[i][TIME % 60];
  } else if (l > ed || r < st)
    return;
  else {
    int m = (l + r) / 2;
    query((2 * i), l, m, st, ed);
    query((2 * i + 1), m + 1, r, st, ed);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, q, l, r;
  int i, j, k;
  char t[10];
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &A[i]);
  build(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    scanf("%s", t);
    scanf("%d%d", &l, &r);
    if (t[0] == 'C') {
      update(1, 1, n, l, r);
    } else {
      TIME = 0;
      query(1, 1, n, l, r - 1);
      printf("%d\n", TIME);
    }
  }
  return 0;
}
