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 T, typename U>
inline void swap(T &a, U &b) {
  a = a ^ b;
  b = a ^ b;
  a = a ^ b;
}
inline void swap(int &a, int &b) {
  a = a ^ b;
  b = a ^ b;
  a = a ^ b;
}
inline void swap(long long &a, long long &b) {
  a = a ^ b;
  b = a ^ b;
  a = a ^ b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (b > a) a = b;
}
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (b < a) a = b;
}
inline int gcd(int a, int b) {
  if (a < b) swap(a, b);
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
inline long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long nchose(int a, int b) {
  long long ans = 1;
  smax(a, b - a);
  int p = 2;
  for (int i = (b - a + 1); i <= (b); i++) {
    ans *= (long long)i;
    while (p <= a && ans % p == 0) ans /= p++;
  }
  return ans;
}
template <typename T, typename U, U (*M)(T), void (*R)(U &, U &)>
class map_reduce {
  vector<U> data;
  int n;

 public:
  map_reduce(vector<T> v) {
    n = 1;
    while (n < v.size()) n *= 2;
    data.resize(n + v.size());
    transform(v.begin(), v.end(), data.begin() + n, M);
    for (int i = data.size() - (v.size() + 1) % 2; i > 2; i -= 2)
      data[i / 2] = data[i - 1], R(data[i / 2], data[i]);
  }
  void put(T a, int p) {
    p += n;
    data[p] = M(a);
    while (p > 1)
      data[p / 2] = data[p - p % 2], R(data[p / 2], data[p + (p + 1) % 2]),
               p /= 2;
  }
  U query(int l, int r) {
    if (l == r) return data[n + l];
    stack<int> q;
    l += n, r += n;
    q.push(r);
    U le = data[l];
    for (; l / 2 != r / 2; l /= 2, r /= 2) {
      if (l % 2 == 0) R(le, data[l + 1]);
      if (r % 2 == 1) q.push(r - 1);
    }
    for (; !q.empty(); q.pop()) R(le, data[q.top()]);
    return le;
  }
  T query1(int l, int r, T init, void (*red)(T &, U &)) {
    if (l == r) return red(init, data[n + l]), init;
    stack<int> q;
    l += n, r += n;
    q.push(r);
    T le = init;
    red(le, data[l]);
    for (; l / 2 != r / 2; l /= 2, r /= 2) {
      if (l % 2 == 0) red(le, data[l + 1]);
      if (r % 2 == 1) q.push(r - 1);
    }
    for (; !q.empty(); q.pop()) red(le, data[q.top()]);
    return le;
  }
  T query2(int l, int r, T init, void (*red)(T &, U &)) {
    l += n, r += n;
    stack<int> q;
    q.push(r);
    red(init, data[l]);
    if (l == r) return init;
    for (; l / 2 != r / 2; l /= 2, r /= 2) {
      if (l % 2 == 0) red(init, data[l + 1]);
      if (r % 2 == 1) q.push(r - 1);
    }
    for (; !q.empty(); q.pop()) red(init, data[q.top()]);
    return init;
  }
};
static long long MOD = 1000000007;
struct st {
  int v[60];
};
inline st mmap(int a) {
  st ret;
  for (int i = (0); i < (60); i++) {
    ret.v[i] = 1 + (i % a == 0);
  }
  return ret;
}
inline void rreduce(st &a, st &b) {
  for (int i = (0); i < (60); i++) {
    a.v[i] += b.v[(i + a.v[i]) % 60];
  }
}
inline void rrreduce(int &a, st &b) { a += b.v[a % 60]; }
int main() {
  vector<int> v;
  int n, q;
  cin >> n;
  v.resize(n);
  for (int &i : v) cin >> i;
  map_reduce<int, st, mmap, rreduce> mm(v);
  cin >> q;
  for (int i = (0); i < (q); i++) {
    char t;
    cin >> t;
    int x, y;
    cin >> x >> y;
    ;
    if (t == 'C') {
      mm.put(y, x - 1);
    } else {
      cout << mm.query2(x - 1, y - 2, 0, rrreduce) << endl;
    }
  }
}
