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 = 1e5 + 1;
int n, q, a[MaxN];
int IT[60][MaxN << 2];
void Enter() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
}
void Modify(int node, int L, int R, int pos) {
  if (L == R) {
    for (int t = 0; t < 60; t++) IT[t][node] = (t % a[pos]) ? 1 : 2;
    return;
  }
  int mid = (L + R) >> 1;
  if (mid < pos) Modify(node << 1 | 1, mid + 1, R, pos);
  if (mid > pos - 1) Modify(node << 1, L, mid, pos);
  for (int t = 0; t < 60; t++) {
    int time = IT[t][node << 1];
    IT[t][node] = time + IT[(time + t) % 60][node << 1 | 1];
  }
}
int Request(int node, int L, int R, int t, int _L, int _R) {
  if (_L <= L && R <= _R) return IT[t][node];
  int mid = (L + R) >> 1;
  if (mid < _L) return Request(node << 1 | 1, mid + 1, R, t, _L, _R);
  if (mid > _R - 1) return Request(node << 1, L, mid, t, _L, _R);
  int time = Request(node << 1, L, mid, t, _L, _R);
  return time + Request(node << 1 | 1, mid + 1, R, (t + time) % 60, _L, _R);
}
void Solve() {
  string c;
  int x, y;
  cin >> q;
  for (int i = 1; i <= n; i++) Modify(1, 1, n, i);
  for (int i = 1; i <= q; i++) {
    cin >> c >> x >> y;
    if (c[0] == 'A')
      cout << Request(1, 1, n, 0, x, y - 1) << endl;
    else
      a[x] = y, Modify(1, 1, n, x);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  Enter();
  Solve();
  return 0;
}
