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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long int mod1 = 60;
int tree[61][100005 * 3];
void update(long long int n, long long int b, long long int e, long long int x,
            long long int val) {
  long long int mid = (b + e) >> 1, l = n * 2, r = l + 1, i;
  if (b == e) {
    for (i = 0; i < mod1; i++) {
      tree[i][n] = (i % val == 0) + 1;
    }
    return;
  }
  if (x <= mid)
    update(l, b, mid, x, val);
  else
    update(r, mid + 1, e, x, val);
  for (i = 0; i < mod1; i++) {
    long long int p = tree[i][l];
    long long int p1 = tree[(i + p) % mod1][r];
    tree[i][n] = p + p1;
  }
}
long long int query(long long int n, long long int b, long long int e,
                    long long int x, long long int y, long long int t) {
  if (b > y || e < x) return 0;
  if (b >= x && e <= y) return tree[t][n];
  long long int mid = (b + e) >> 1, l = n * 2, r = l + 1;
  long long int p = query(l, b, mid, x, y, t);
  long long int p1 = query(r, mid + 1, e, x, y, (t + p) % mod1);
  return p + p1;
}
int main() {
  int n, i, j, q, x, y;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    update(1, 1, n, i, x);
  }
  scanf("%d", &q);
  getchar();
  while (q--) {
    char ch;
    scanf("%c", &ch);
    scanf("%d%d", &x, &y);
    getchar();
    if (ch == 'A') {
      int ans = query(1, 1, n, x, y - 1, 0);
      printf("%d\n", ans);
    } else
      update(1, 1, n, x, y);
  }
}
