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 calc(const vector<int>& a, int L, int R, int t0) {
  ;
  int t = t0;
  for (int i = L; i != R; ++i) {
    assert(i < R);
    if (t % a[i] == 0) {
      ++t;
    }
    ++t;
  };
  return t - t0;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  const int MOD = 60;
  const int k = sqrt(MOD * n + 0.0);
  const int m = (n + k - 1) / k + 1;
  ;
  int q;
  scanf("%d", &q);
  vector<vector<int> > precalc(m, vector<int>(MOD, -1));
  for (int i = 0; i < q; ++i) {
    char ch;
    int x, y;
    scanf(" %c%d%d", &ch, &x, &y);
    ;
    if (ch == 'A') {
      --x, --y;
      int t = 0;
      for (int i = x; i != y;) {
        ;
        assert(i < y);
        if (i % k == 0 && i + k <= y) {
          int& val = precalc[i / k][t % MOD];
          if (val == -1) {
            val = calc(a, i, i + k, t % MOD);
          }
          t += val;
          i += k;
        } else {
          if (t % a[i] == 0) {
            ++t;
          }
          ++t;
          ++i;
        }
      }
      printf("%d\n", t);
    } else if (ch == 'C') {
      --x;
      if (a[x] != y) {
        a[x] = y;
        precalc[x / k].assign(MOD, -1);
      }
    } else {
      throw 42;
    }
  }
  return 0;
}
