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;
template <typename M>
struct SegTree {
  using Op = function<M(const M&, const M&)>;
  M ID;
  Op op;
  int N;
  vector<M> a;
  SegTree(int n, M leaf_default, M _ID, Op _op) : ID(_ID), op(_op) {
    init_space(n);
    fill(a.begin() + N, a.begin() + N + n, leaf_default);
    build();
  }
  SegTree(const vector<M> leaves, M _ID, Op _op) : ID(_ID), op(_op) {
    int n = leaves.size();
    init_space(n);
    copy(leaves.begin(), leaves.end(), a.begin() + N);
    build();
  }
  void init_space(int n) {
    N = 1;
    while (N < n) N <<= 1;
    a.assign(N << 1, ID);
  }
  inline void pull(int i) { a[i] = op(a[i << 1], a[i << 1 | 1]); }
  void build() {
    for (int i = N - 1; i >= 1; i--) pull(i);
  }
  void assign(int p, M val) {
    assert(0 <= p && p < N);
    for (a[p += N] = val; p >>= 1;) pull(p);
  }
  M query(int l, int r) const {
    assert(0 <= l && r <= N);
    M resl(ID), resr(ID);
    for (l += N, r += N; l < r; l >>= 1, r >>= 1) {
      if (l & 1) resl = op(resl, a[l++]);
      if (r & 1) resr = op(a[--r], resr);
    }
    return op(resl, resr);
  }
  M query_point(int p) const {
    assert(0 <= p && p < N);
    return a[p + N];
  }
  M query_all() const { return a[1]; }
};
const int P = 60;
void solve() {
  int n;
  cin >> n;
  using M = array<int, P>;
  auto create = [&](int mod) {
    M r = {};
    for (int t = 0; t < P; t++) {
      r[t] = 1 + (t % mod == 0);
    }
    return r;
  };
  M B[7];
  for (int i = 2; i <= 6; i++) {
    B[i] = create(i);
  }
  vector<M> a(n);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a[i] = B[x];
  }
  SegTree<M> s(a, M{}, [](M u, M v) {
    M r = {};
    for (int t = 0; t < P; t++) {
      int dt = u[t];
      r[t] = dt + v[(t + dt) % P];
    }
    return r;
  });
  int q;
  cin >> q;
  while (q--) {
    char op;
    int x, y;
    cin >> op >> x >> y;
    if (op == 'A') {
      x--;
      y--;
      cout << s.query(x, y)[0] << '\n';
    } else {
      x--;
      s.assign(x, B[y]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
