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 amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
const int K = 60;
struct Sum {
  char func[K];
  short time[K];
  Sum() {
    for (int(k) = 0; (k) < (int)(K - 1); ++(k)) func[k] = k + 1;
    for (int(k) = 0; (k) < (int)(K - 1); ++(k)) time[k] = 0;
    func[K - 1] = 0;
    time[K - 1] = 1;
  }
  explicit Sum(int a) {
    for (int(k) = 0; (k) < (int)(K - 1); ++(k)) func[k] = k + 1;
    for (int(k) = 0; (k) < (int)(K - 1); ++(k)) time[k] = 0;
    func[K - 1] = 0;
    time[K - 1] = 1;
    for (int k = 0; k < K; k += a) func[k] = k + 2;
    if (a == 2) {
      func[K - 2] = 0;
      time[K - 2] = 1;
    }
  }
  Sum operator+(const Sum &that) const {
    Sum t;
    for (int(k) = 0; (k) < (int)(K); ++(k)) {
      int l = func[k];
      t.func[k] = that.func[l];
      t.time[k] = time[k] + that.time[l];
    }
    return t;
  }
  Sum &operator+=(const Sum &that) { return *this = *this + that; }
};
struct GetRangeSegmentTree {
  static Sum combineVal(const Sum &x, const Sum &y) { return x + y; }
  static void combinesVal(Sum &x, const Sum &y) { x += y; }
  static Sum identityVal() { return Sum(); }
  vector<Sum> nodes;
  int n;
  void init(int n_, const Sum &v = Sum()) { init(vector<Sum>(n_, v)); }
  void init(const vector<Sum> &u) {
    n = 1;
    while (n < (int)u.size()) n *= 2;
    nodes.resize(n, identityVal());
    nodes.insert(nodes.end(), u.begin(), u.end());
    nodes.resize(n * 2, identityVal());
    for (int i = n - 1; i > 0; --i)
      nodes[i] = combineVal(nodes[i * 2], nodes[i * 2 + 1]);
  }
  Sum get(int i) { return nodes[i + n]; }
  Sum getRange(int l, int r) const {
    Sum m = identityVal();
    int indices[64];
    int k = 0;
    for (; l && l + (l & -l) <= r; l += l & -l)
      combinesVal(m, nodes[(n + l) / (l & -l)]);
    for (; l < r; r -= r & -r) indices[k++] = (n + r) / (r & -r) - 1;
    while (--k >= 0) combinesVal(m, nodes[indices[k]]);
    return m;
  }
  void set(int i, const Sum &x) {
    i += n;
    nodes[i] = x;
    for (i >>= 1; i > 0; i >>= 1)
      nodes[i] = combineVal(nodes[i * 2], nodes[i * 2 + 1]);
  }
};
int main() {
  int n;
  scanf("%d", &n);
  vector<Sum> vals(7);
  for (int(k) = (int)(2); (k) <= (int)(6); ++(k)) vals[k] = Sum(k);
  vector<Sum> initval(n);
  for (int(i) = 0; (i) < (int)(n); ++(i)) {
    int a;
    scanf("%d", &a);
    initval[i] = vals[a];
  }
  GetRangeSegmentTree segt;
  segt.init(initval);
  int q;
  scanf("%d", &q);
  char ty[2];
  for (int(ii) = 0; (ii) < (int)(q); ++(ii)) {
    int x, y;
    scanf("%s%d%d", ty, &x, &y);
    if (*ty == 'A') {
      --x, --y;
      Sum sum = segt.getRange(x, y);
      int ans = (int)sum.time[0] * K + sum.func[0];
      printf("%d\n", ans);
    } else if (*ty == 'C') {
      --x;
      segt.set(x, vals[y]);
    }
  }
  return 0;
}
