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;
struct MagicTree {
  vector<vector<int> > data;
  vector<int> a;
  int n;
  MagicTree() {}
  MagicTree(int a[], int n) {
    this->a = vector<int>(a, a + n);
    this->n = n;
    data.assign(4 * n, vector<int>(60));
    build(0, 0, n);
  }
  void build(int k, int l, int r) {
    if (l + 1 == r) {
      for (int t = 0; t < 60; ++t) {
        data[k][t] = t % a[l] == 0 ? 2 : 1;
      }
    } else {
      build(2 * k + 1, l, (l + r) / 2);
      build(2 * k + 2, (l + r) / 2, r);
      for (int t = 0; t < 60; ++t) {
        data[k][t] =
            data[2 * k + 1][t] + data[2 * k + 2][(t + data[2 * k + 1][t]) % 60];
      }
    }
  }
  void update(int x, int y) {
    a[x] = y;
    update(0, 0, n, x);
  }
  void update(int k, int l, int r, int x) {
    if (x < l || x >= r) {
      return;
    }
    if (l + 1 == r) {
      for (int t = 0; t < 60; ++t) {
        data[k][t] = t % a[x] == 0 ? 2 : 1;
      }
    } else {
      update(2 * k + 1, l, (l + r) / 2, x);
      update(2 * k + 2, (l + r) / 2, r, x);
      for (int t = 0; t < 60; ++t) {
        data[k][t] =
            data[2 * k + 1][t] + data[2 * k + 2][(t + data[2 * k + 1][t]) % 60];
      }
    }
  }
  int get(int l, int r) { return get(0, 0, n, l, r, 0); }
  int get(int k, int l, int r, int x, int y, int t) {
    if (y <= l || r <= x || l >= r) {
      return 0;
    }
    if (x <= l && r <= y) {
      return data[k][t];
    } else {
      int tmp = get(2 * k + 1, l, (l + r) / 2, x, y, t);
      return tmp + get(2 * k + 2, (l + r) / 2, r, x, y, (t + tmp) % 60);
    }
  }
};
const int N = (int)1e5;
int a[N];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  int q;
  cin >> q;
  MagicTree tree(a, n);
  for (int i = 0; i < q; ++i) {
    char type;
    cin >> type;
    int x;
    cin >> x;
    --x;
    int y;
    cin >> y;
    if (type == 'A') {
      --y;
      printf("%d\n", tree.get(x, y));
    } else {
      tree.update(x, y);
    }
  }
  return 0;
}
