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;
struct node {
  int l, r, res[65];
} tree[400005];
int n, q, fuck[100005];
void merge(int k) {
  for (int i = 0; i < 60; i++)
    tree[k].res[i] = (tree[k * 2].res[i] +
                      tree[k * 2 + 1].res[(i + tree[k * 2].res[i]) % 60]);
}
void build(int i, int l, int r) {
  tree[i].l = l, tree[i].r = r;
  if (l == r) {
    for (int j = 0; j < 60; j++) tree[i].res[j] = 1 + (j % fuck[l] == 0);
    return;
  }
  int mid = (l + r) / 2;
  build(i * 2, l, mid);
  build(i * 2 + 1, mid + 1, r);
  merge(i);
}
void update(int i, int l, int x) {
  if (tree[i].l == tree[i].r) {
    for (int j = 0; j < 60; j++) tree[i].res[j] = 1 + (j % x == 0);
    return;
  }
  int mid = (tree[i].l + tree[i].r) / 2;
  if (l <= mid)
    update(i * 2, l, x);
  else
    update(i * 2 + 1, l, x);
  merge(i);
}
int query(int i, int l, int r, int ind) {
  if (l <= tree[i].l && tree[i].r <= r) return tree[i].res[ind];
  if (tree[i].l > r || tree[i].r < l) return 0;
  int ret = query(i * 2, l, r, ind);
  return (ret + query(i * 2 + 1, l, r, (ind + ret) % 60));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &fuck[i]);
  build(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    char c[5];
    int x, y;
    scanf("%s%d%d", c, &x, &y);
    if (c[0] == 'C')
      update(1, x, y);
    else
      printf("%d\n", query(1, x, y - 1, 0));
  }
}
