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 P = 60;
struct SegTree {
  int n;
  vector<array<int, P> > t;
  SegTree(int n_) {
    n = 1;
    while (n < n_) n <<= 1;
    t.resize(2 * n - 1);
  }
  array<int, P> calc(const array<int, P> &u, const array<int, P> &v) {
    array<int, P> w = {0};
    for (int i = 0; i < P; i++) {
      w[i] = v[(i + u[i]) % P] + u[i];
    }
    return w;
  }
  void update(int k, const array<int, P> &v) {
    k += n - 1;
    t[k] = v;
    while (k > 0) {
      k = (k - 1) >> 1;
      t[k] = calc(t[2 * k + 1], t[2 * k + 2]);
    }
  }
  array<int, P> query(int a, int b, int k, int l, int r) {
    if (b <= l || r <= a) {
      return array<int, P>{0};
    } else if (a <= l && r <= b) {
      return t[k];
    } else {
      return calc(query(a, b, 2 * k + 1, l, (l + r) / 2),
                  query(a, b, 2 * k + 2, (l + r) / 2, r));
    }
  }
  array<int, P> query(int a, int b) { return query(a, b, 0, 0, n); }
};
int n, q;
int main() {
  vector<array<int, P> > vs(7);
  for (int i = 2; i <= 6; i++) {
    for (int j = 0; j < P; j++) {
      int k = 0;
      while ((j + k) % i == 0) {
        k++;
      }
      vs[i][j] = k + 1;
    }
  }
  cin >> n;
  SegTree st(n);
  for (int i = 0; i < n; i++) {
    int v;
    cin >> v;
    st.update(i, vs[v]);
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    char c;
    scanf(" %c", &c);
    if (c == 'C') {
      int k, v;
      scanf("%d%d", &k, &v);
      k--;
      st.update(k, vs[v]);
    } else if (c == 'A') {
      int s, t;
      scanf("%d%d", &s, &t);
      s--, t--;
      printf("%d\n", st.query(s, t)[0]);
    } else {
      assert(false);
    }
  }
}
