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 <class C>
void mini(C& a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
  a4 = max(a4, b4);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
  return out << "(" << pair.first << ", " << pair.second << ")";
}
template <class A, class B, class C>
struct Triple {
  A first;
  B second;
  C third;
};
template <class T>
void ResizeVec(T&, vector<int>) {}
template <class T>
void ResizeVec(vector<T>& vec, vector<int> sz) {
  vec.resize(sz[0]);
  sz.erase(sz.begin());
  if (sz.empty()) {
    return;
  }
  for (T& v : vec) {
    ResizeVec(v, sz);
  }
}
const int R = 60;
void Join(vector<int>& v1, vector<int>& v2, vector<int>& res) {
  for (int i = 0; i < (R); ++i) {
    res[i] = v1[i] + v2[(i + v1[i]) % R];
  }
}
const int M = (1 << 17);
vector<int> drz[2 * M + 5];
void AssignVec(int a, int per) {
  a += M - 1;
  for (int i = 0; i < (R); ++i) {
    drz[a][i] = 1 + (i % per == 0);
  }
}
void Change(int a, int per) {
  AssignVec(a, per);
  a += M - 1;
  a /= 2;
  while (a) {
    Join(drz[2 * a], drz[2 * a + 1], drz[a]);
    a /= 2;
  }
}
int Read(int a, int b) {
  a += M - 1;
  b += M - 1;
  int act_t = drz[a][0];
  if (a == b) {
    return act_t;
  }
  vector<int> from_left_inds;
  vector<int> from_right_inds;
  from_right_inds.push_back(b);
  while (a / 2 != b / 2) {
    if (a % 2 == 0) {
      from_left_inds.push_back(a + 1);
    }
    if (b % 2 == 1) {
      from_right_inds.push_back(b - 1);
    }
    a /= 2;
    b /= 2;
  }
  for (auto x : from_left_inds) {
    act_t += drz[x][act_t % R];
  }
  reverse((from_right_inds).begin(), (from_right_inds).end());
  for (auto x : from_right_inds) {
    act_t += drz[x][act_t % R];
  }
  return act_t;
}
int main() {
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  double beg_clock = 1.0 * clock() / CLOCKS_PER_SEC;
  for (long long i = (1); i <= (2 * M); ++i) {
    drz[i].resize(R);
  }
  long long n;
  cin >> n;
  ;
  for (long long i = (1); i <= (n); ++i) {
    long long per;
    cin >> per;
    ;
    AssignVec(i, per);
  }
  for (long long i = M - 1; i; i--) {
    Join(drz[2 * i], drz[2 * i + 1], drz[i]);
  }
  long long q;
  cin >> q;
  ;
  for (long long i = (1); i <= (q); ++i) {
    char c;
    cin >> c;
    ;
    long long x, y;
    cin >> x >> y;
    ;
    if (c == 'C') {
      Change(x, y);
    } else {
      cout << Read(x, y - 1) << "\n";
    }
  }
  return 0;
}
