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 A[1 << 17];
int R[2 << 17][60];
int I[17 + 1];
void update(int level, int pos) {
  int i;
  if (level == 0) {
    for (i = (0); i < (60); ++i)
      R[I[level] + pos][i] = (A[pos] != 0 && i % A[pos] == 0) ? 2 : 1;
  } else {
    for (i = (0); i < (60); ++i) {
      int r = R[I[level - 1] + pos + pos][i];
      R[I[level] + pos][i] = r + R[I[level - 1] + pos + pos + 1][(i + r) % 60];
    }
  }
}
int get(int level, int pos, int cycle, int x, int y) {
  int from = pos << level;
  int to = (pos + 1) << level;
  if (from >= y || to <= x) return 0;
  if (x <= from && to <= y) return R[I[level] + pos][cycle];
  int r = get(level - 1, pos + pos, cycle, x, y);
  return r + get(level - 1, pos + pos + 1, (cycle + r) % 60, x, y);
}
int main() {
  int n;
  scanf("%d", &n);
  int i, j;
  memset(A, 1, sizeof(A));
  for (i = (0); i < (n); ++i) scanf("%d", &A[i]);
  I[0] = 0;
  for (i = (0); i < (17); ++i) I[i + 1] = I[i] + (1 << (17 - i));
  for (i = (0); i < (17 + 1); ++i)
    for (j = (0); j < ((1 << (17 - i))); ++j) update(i, j);
  int q;
  scanf("%d", &q);
  for (i = (0); i < (q); ++i) {
    char buf[4];
    int x, y;
    scanf("%s%d%d", buf, &x, &y);
    if (buf[0] == 'A') {
      --x;
      --y;
      printf("%d\n", get(17, 0, 0, x, y));
    } else {
      --x;
      A[x] = y;
      for (j = (0); j < (17 + 1); ++j) update(j, x >> j);
    }
  }
  return 0;
};
