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 n, m;
int a[100001];
int tree[60][4 * 100001];
int query(int cx, int cy, int qx, int qy, int res, int pos) {
  if (qy < cx || cy < qx) return res;
  if (qx <= cx && cy <= qy) return res + tree[res % 60][pos];
  int mid = (cx + cy) >> 1;
  res = query(cx, mid, qx, qy, res, pos * 2);
  return query(mid + 1, cy, qx, qy, res, pos * 2 + 1);
}
void update(int cx, int cy, int q, int val, int pos) {
  if (cy < q || q < cx) return;
  if (cx == cy) {
    for (int kd = 0; kd < 60; kd++) tree[kd][pos] = 1 + !(kd % val);
    return;
  }
  int mid = (cx + cy) >> 1;
  update(cx, mid, q, val, pos * 2);
  update(mid + 1, cy, q, val, pos * 2 + 1);
  for (int kd = 0; kd < 60; kd++) {
    int tmp = tree[kd][pos * 2];
    tree[kd][pos] = tmp + tree[(kd + tmp) % 60][pos * 2 + 1];
  }
}
void init(int x, int y, int pos) {
  if (x == y) {
    for (int kd = 0; kd < 60; kd++) tree[kd][pos] = 1 + !(kd % a[x]);
    return;
  }
  int mid = (x + y) >> 1;
  init(x, mid, pos * 2);
  init(mid + 1, y, pos * 2 + 1);
  for (int kd = 0; kd < 60; kd++) {
    int tmp = tree[kd][pos * 2];
    tree[kd][pos] = tmp + tree[(kd + tmp) % 60][pos * 2 + 1];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  init(1, n, 1);
  scanf("%d", &m);
  for (int i = 0, x, y; i < m; i++) {
    char s[5];
    scanf("%s%d%d", s, &x, &y);
    if (s[0] == 'C')
      update(1, n, x, y, 1);
    else
      printf("%d\n", query(1, n, x, y - 1, 0, 1));
  }
  return 0;
}
