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>
#pragma GCC optimize("-O3")
using namespace std;
void _main();
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  _main();
}
template <class V, int NV>
struct SegTree {
  inline V comp(V& l, V& r) {
    auto v = vector<int>(60);
    for (int i = 0; i < 60; i++) v[i] = l[i] + r[(i + l[i]) % 60];
    return v;
  };
  vector<V> val;
  SegTree() { val = vector<V>(NV * 2, vector<int>(60)); }
  V get(int x, int y, int l = 0, int r = NV, int k = 1) {
    if (r <= x || y <= l) return vector<int>(60);
    if (x <= l && r <= y) return val[k];
    auto A = get(x, y, l, (l + r) / 2, k * 2);
    auto B = get(x, y, (l + r) / 2, r, k * 2 + 1);
    return comp(A, B);
  }
  void update(int i, V v) {
    i += NV;
    val[i] = v;
    while (i > 1) i >>= 1, val[i] = comp(val[i * 2], val[i * 2 + 1]);
  }
};
int N, A[101010], M;
SegTree<vector<int>, 1 << 17> st;
void update(int i, int a) {
  auto v = vector<int>(60);
  for (int j = 0; j < 60; j++) {
    if ((j % a) == 0)
      v[j] = 2;
    else
      v[j] = 1;
  }
  st.update(i, v);
}
void _main() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> A[i];
  for (int i = 0; i < N; i++) update(i, A[i]);
  cin >> M;
  for (int i = 0; i < M; i++) {
    string s;
    int x, y;
    cin >> s >> x >> y;
    if (s == "C") {
      x--;
      A[x] = y;
      update(x, A[x]);
    } else {
      x--;
      y--;
      auto v = st.get(x, y);
      printf("%d\n", v[0]);
    }
  }
}
