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;
struct STree {
  vector<vector<pair<int, short>>> st;
  int n;
  STree(int n) : st(4 * n + 5, vector<pair<int, short>>(60, {-1, -1})), n(n) {}
  void oper(vector<pair<int, short>> &a, vector<pair<int, short>> &b,
            vector<pair<int, short>> &c) {
    for (int i = 0; i < 60; i++)
      c[i] = {a[i].first + b[a[i].second].first, b[a[i].second].second};
  }
  void init(int k, int s, int e, vector<int> &a) {
    if (s + 1 == e) {
      for (int i = 0; i < 60; i++) {
        if (i % a[s] == 0)
          st[k][i] = {2, (i + 2) % 60};
        else
          st[k][i] = {1, (i + 1) % 60};
      }
      return;
    }
    int m = (s + e) / 2;
    init(2 * k, s, m, a);
    init(2 * k + 1, m, e, a);
    oper(st[2 * k], st[2 * k + 1], st[k]);
  }
  void upd(int k, int s, int e, int p, int v) {
    if (s + 1 == e) {
      for (int i = 0; i < 60; i++) {
        if (i % v == 0)
          st[k][i] = {2, (i + 2) % 60};
        else
          st[k][i] = {1, (i + 1) % 60};
      }
      return;
    }
    int m = (s + e) / 2;
    if (p < m)
      upd(2 * k, s, m, p, v);
    else
      upd(2 * k + 1, m, e, p, v);
    oper(st[2 * k], st[2 * k + 1], st[k]);
  }
  pair<int, short> query(int k, int s, int e, int a, int b, int mod = 0) {
    if (s >= b || e <= a) return {-1, -1};
    if (s >= a && e <= b) return st[k][mod];
    int m = (s + e) / 2;
    pair<int, short> res = {0, mod};
    auto izq = query(2 * k, s, m, a, b, mod);
    if (izq.first != -1) res.first += izq.first, res.second = izq.second;
    auto der = query(2 * k + 1, m, e, a, b, res.second);
    if (der.first != -1) res.first += der.first, res.second = der.second;
    return res;
  }
  void init(vector<int> &a) { init(1, 0, n, a); }
  void upd(int p, int v) { upd(1, 0, n, p, v); }
  pair<int, short> query(int a, int b) { return query(1, 0, n, a, b); }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  STree rmq(n);
  rmq.init(a);
  int q;
  cin >> q;
  while (q--) {
    char c;
    int x, y;
    cin >> c >> x >> y;
    if (c == 'C')
      rmq.upd(x - 1, y);
    else
      cout << rmq.query(x - 1, y - 1).first << '\n';
  }
  return 0;
}
