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 MAX = (1 << 17);
int seg[2 * MAX + 10][65];
void propagate_up(int p) {
  if (p >= MAX) return;
  for (int i = 0; i < 60; i++) {
    int f = seg[p * 2][i];
    seg[p][i] = f + seg[p * 2 + 1][(i + f) % 60];
  }
}
void insert(int a, int l) {
  int p = l + MAX;
  for (int i = 0; i < 60; i++) {
    if (i % a == 0)
      seg[p][i] = 2;
    else
      seg[p][i] = 1;
  }
  for (; p > 0; p /= 2) propagate_up(p);
}
int lookup2(int l, int r, int p, int a, int b, int t) {
  if (a >= r || b <= l) return 0;
  if (l <= a && b <= r) return seg[p][t % 60];
  int m = (a + b) / 2;
  int f = lookup2(l, r, p * 2, a, m, t);
  return f + lookup2(l, r, p * 2 + 1, m, b, t + f);
}
int lookup(int l, int r) { return lookup2(l, r, 1, 0, MAX, 0); }
int N;
int A[MAX];
int main() {
  ios::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    insert(A[i], i);
  }
  int Q;
  cin >> Q;
  while (Q--) {
    char t;
    int a, b;
    cin >> t >> a >> b;
    if (t == 'A')
      cout << lookup(a, b) << '\n';
    else
      insert(b, a);
  }
}
