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 = 1e5 + 10;
struct Node {
  int p[60];
  Node() {
    for (int i = 0; i < 60; i++) p[i] = 0;
  }
  Node(int a) {
    for (int i = 0; i < 60; i++) p[i] = 1 + ((i % a) == 0);
  }
};
Node merge(Node a, Node b) {
  if (a.p[0] == 0) return b;
  if (b.p[0] == 0) return a;
  Node res;
  for (int i = 0; i < 60; i++) res.p[i] = a.p[i] + b.p[(a.p[i] + i) % 60];
  return res;
}
int A[Max];
Node seg[Max * 4];
int n;
void build(int tl = 1, int tr = n + 1, int idx = 1) {
  if (tr - tl == 1) {
    seg[idx] = {A[tl]};
    return;
  }
  int mid = (tl + tr) / 2;
  build(tl, mid, idx * 2);
  build(mid, tr, idx * 2 + 1);
  seg[idx] = merge(seg[idx * 2], seg[idx * 2 + 1]);
}
void upd(int i, int x, int tl = 1, int tr = n + 1, int idx = 1) {
  if (tr - tl == 1) {
    seg[idx] = {x};
    return;
  }
  int mid = (tl + tr) / 2;
  if (i < mid)
    upd(i, x, tl, mid, idx * 2);
  else
    upd(i, x, mid, tr, idx * 2 + 1);
  seg[idx] = merge(seg[idx * 2], seg[idx * 2 + 1]);
}
Node get(int l, int r, int tl = 1, int tr = n + 1, int idx = 1) {
  if (l <= tl && tr <= r) return seg[idx];
  if (r <= tl || tr <= l) return {};
  int mid = (tl + tr) / 2;
  return merge(get(l, r, tl, mid, idx * 2), get(l, r, mid, tr, idx * 2 + 1));
}
int main() {
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> A[i];
  build();
  int q;
  cin >> q;
  while (q--) {
    char ty;
    cin >> ty;
    int l, r;
    cin >> l >> r;
    if (ty == 'C') {
      upd(l, r);
    } else {
      Node ans = get(l, r);
      cout << (ans.p[1] + 1) << '\n';
    }
  }
}
