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;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
const double PI = acos(-1);
int seg[300000][60], arr[100005];
void calc(int v) {
  static int temp, l, r;
  l = v << 1;
  r = v << 1 | 1;
  for (int i = 0; i < 60; i++) {
    temp = seg[l][i];
    seg[v][i] = temp + seg[r][(temp + i) % 60];
  }
}
void build(int v, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 60; i++)
      if (i % arr[l])
        seg[v][i] = 1;
      else
        seg[v][i] = 2;
    return;
  }
  int mid = (l + r) / 2;
  build(v << 1, l, mid);
  build(v << 1 | 1, mid + 1, r);
  calc(v);
}
void upd(int v, int l, int r, int q, int val) {
  if (l == r) {
    arr[l] = val;
    for (int i = 0; i < 60; i++)
      if (i % arr[l])
        seg[v][i] = 1;
      else
        seg[v][i] = 2;
    return;
  }
  int mid = (l + r) / 2;
  if (q <= mid)
    upd(v << 1, l, mid, q, val);
  else
    upd(v << 1 | 1, mid + 1, r, q, val);
  calc(v);
}
void query(int v, int l, int r, int ql, int qr, int& time) {
  if (ql <= l and r <= qr) {
    time += seg[v][time % 60];
    return;
  }
  int mid = (l + r) / 2;
  if (ql <= mid) query(v << 1, l, mid, ql, qr, time);
  if (qr > mid) query(v << 1 | 1, mid + 1, r, ql, qr, time);
}
int tejas_919(int kkkk) {
  lli n, m, k, q, u, v, temp = 0, ans = 0;
  cin >> n;
  for (int i = 1; i < n + 1; i++) cin >> arr[i];
  build(1, 1, n);
  cin >> q;
  for (int i = 0; i < q; i++) {
    char ch;
    cin >> ch >> u >> v;
    if (ch == 'A') {
      int time = 0;
      query(1, 1, n, u, v - 1, time);
      cout << time << '\n';
    } else
      upd(1, 1, n, u, v);
  }
  return 0;
}
signed main() {
  if (!0) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  cout << fixed << setprecision(10);
  int t = 1;
  for (int i = 0; i < t; i++) {
    tejas_919(i + 1);
  }
  {};
}
