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>
char qt[5];
int a[100010];
const int N = 1 << 18;
int b[N + N][60];
inline void go(int x, int &t) {
  if (t % a[x] == 0) t++;
  t++;
}
void upd(int i, int y) {
  a[i] = y;
  for (int j = 0; j < 60; j++) {
    b[i + N][j] = j % a[i] ? 1 : 2;
  }
  i += N;
  while (i > 1) {
    i >>= 1;
    for (int j = 0; j < 60; j++) {
      int t = b[i + i][j] + j;
      b[i][j] = t + b[i + i + 1][t % 60] - j;
    }
  }
}
int query(int l, int r, int L = 0, int R = N, int p = 1, int t = 0) {
  if (r <= L || R <= l) return t;
  if (l <= L && R <= r) {
    return t + b[p][t % 60];
  }
  int mid = (L + R) / 2;
  t = query(l, r, L, mid, p + p, t);
  t = query(l, r, mid, R, p + p + 1, t);
  return t;
}
int main() {
  int n, q, x, y;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 60; j++) {
      b[i + N][j] = j % a[i] ? 1 : 2;
    }
  }
  for (int i = N - 1; i > 0; i--) {
    for (int j = 0; j < 60; j++) {
      int t = b[i + i][j] + j;
      b[i][j] = t + b[i + i + 1][t % 60] - j;
    }
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%s%d%d", qt, &x, &y);
    --x;
    if (qt[0] == 'C') {
      upd(x, y);
    } else {
      --y;
      printf("%d\n", query(x, y));
    }
  }
  return 0;
}
