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 maxn = 100010;
int n, Q;
int A[maxn];
int sm[60][maxn << 2];
char st[10];
void update(int i, int l, int r, int x) {
  if (l == r) {
    for (int j = 0; j < 60; j++) sm[j][i] = 1 + (j % A[l] == 0);
    return;
  }
  int M = l + r >> 1;
  if (x <= M)
    update(i << 1, l, M, x);
  else
    update(i << 1 | 1, M + 1, r, x);
  for (int j = 0; j < 60; j++)
    sm[j][i] = sm[j][i << 1] + sm[(j + sm[j][i << 1]) % 60][i << 1 | 1];
}
int query(int i, int l, int r, int L, int R, int tim) {
  if (L <= l & r <= R) return sm[tim][i];
  int M = l + r >> 1, ans = 0;
  if (L <= M) ans += query(i << 1, l, M, L, R, tim);
  if (R > M) ans += query(i << 1 | 1, M + 1, r, L, R, (tim + ans) % 60);
  return ans;
}
void Work() {
  scanf("%d", &Q);
  while (Q--) {
    int x, y;
    scanf("%s%d%d", st, &x, &y);
    if (st[0] == 'A')
      printf("%d\n", query(1, 1, n, x, y - 1, 0));
    else
      A[x] = y, update(1, 1, n, x);
  }
}
void Init() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]), update(1, 1, n, i);
}
int main() {
  Init();
  Work();
  return 0;
}
