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;
const int N = 1e5 + 5;
int tree[N * 4][60];
int a[N];
void init(int k, int m, int n) {
  if (m == n) {
    for (int j = 0; j < 60; ++j)
      if (j % a[m] == 0)
        tree[k][j] = 2;
      else
        tree[k][j] = 1;
    return;
  }
  int mid = m + n >> 1, z1 = k << 1, z2 = z1 + 1;
  init(z1, m, mid);
  init(z2, mid + 1, n);
  for (int j = 0; j < 60; ++j) {
    int p = (j + tree[z1][j]) % 60;
    tree[k][j] = tree[z1][j] + tree[z2][p];
  }
}
void modify(int k, int m, int n, int x, int y) {
  if (m > x || n < x) return;
  if (m == n) {
    for (int j = 0; j < 60; ++j)
      if (j % y == 0)
        tree[k][j] = 2;
      else
        tree[k][j] = 1;
    return;
  }
  int mid = m + n >> 1, z1 = k << 1, z2 = z1 + 1;
  modify(z1, m, mid, x, y);
  modify(z2, mid + 1, n, x, y);
  for (int j = 0; j < 60; ++j) {
    int p = (j + tree[z1][j]) % 60;
    tree[k][j] = tree[z1][j] + tree[z2][p];
  }
}
void query(int k, int m, int n, int l, int r, int &ans) {
  if (m > r || n < l) return;
  if (m >= l && n <= r) {
    ans += tree[k][ans % 60];
    return;
  }
  int mid = m + n >> 1, z1 = k << 1, z2 = z1 + 1;
  query(z1, m, mid, l, r, ans);
  query(z2, mid + 1, n, l, r, ans);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  init(1, 1, n);
  int m;
  scanf("%d", &m);
  while (m--) {
    int ch;
    while (ch = getchar(), !isupper(ch))
      ;
    if (ch == 'C') {
      int x, y;
      scanf("%d%d", &x, &y);
      modify(1, 1, n, x, y);
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      int ans = 0;
      query(1, 1, n, x, y - 1, ans);
      printf("%d\n", ans);
    }
  }
  return 0;
}
