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 inf = 1e9;
const double eps = 1e-9;
const int INF = inf;
const double EPS = eps;
struct __timestamper {};
const int PER = 60;
class segm_tree {
  struct Data {
    int toadd[PER];
    Data() { fill(toadd, toadd + PER, 0); }
    Data(int x) {
      assert(2 <= x && PER % x == 0);
      fill(toadd, toadd + PER, 1);
      for (int i = 0; i < PER; i += x) toadd[i]++;
    }
    int &operator[](int x) { return toadd[x]; }
    int operator[](int x) const { return toadd[x]; }
    friend Data operator+(const Data &a, const Data &b) {
      Data res;
      for (int st = 0; st < PER; st++) {
        res[st] = a[st] + b[(st + a[st]) % PER];
      }
      return res;
    }
  };
  int off;
  vector<Data> tr;

 public:
  segm_tree(vector<int> &as) {
    off = 1;
    while (off < ((int)(as).size())) off <<= 1;
    tr = vector<Data>(2 * off);
    for (int i = 0; i < (((int)(as).size())); i++) tr[off + i] = as[i];
    for (int i = off - 1; i >= 1; i--) tr[i] = tr[2 * i] + tr[2 * i + 1];
  }
  void set(int x, int v) {
    tr[x += off] = v;
    for (x >>= 1; x >= 1; x >>= 1) tr[x] = tr[2 * x] + tr[2 * x + 1];
  }
  int get(int l, int r) {
    Data resl, resr;
    l += off;
    r += off;
    while (l <= r) {
      if (l & 1) resl = resl + tr[l++];
      if (!(r & 1)) resr = tr[r--] + resr;
      l >>= 1;
      r >>= 1;
    }
    Data res = resl + resr;
    return res[0];
  }
};
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    vector<int> as(n);
    for (int i = 0; i < (n); i++) scanf("%d", &as[i]);
    segm_tree tr(as);
    int q;
    scanf("%d", &q);
    while (q-- > 0) {
      char op;
      scanf(" %c", &op);
      if (op == 'A') {
        int l, r;
        scanf("%d%d", &l, &r), l--, r -= 2;
        printf("%d\n", tr.get(l, r));
      } else if (op == 'C') {
        int x, v;
        scanf("%d%d", &x, &v), x--;
        tr.set(x, v);
      } else {
        assert(false);
      }
    }
  }
  return 0;
}
