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>
const int mod = 1000000007;
const int inf = 1000000009;
const long long INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 1e-9;
using namespace std;
int T[100005], D[300005][61];
int last = 0;
int getsum(int a, int b, int lo, int hi, int u) {
  if (b <= a) return 0;
  if (a == lo && b == hi) {
    int x = D[u][last];
    last = (last + x) % 60;
    return x;
  }
  int mid = (lo + hi) / 2;
  int L = getsum(a, min(b, mid), lo, mid, 2 * u);
  int P = getsum(max(a, mid), b, mid, hi, 2 * u + 1);
  return L + P;
}
void update(int u, int c) {
  for (int j = 0; j < 60; j++) {
    D[u][j] = 1;
    if (j % c == 0) D[u][j]++;
  }
  while (u != 1) {
    u /= 2;
    for (int j = 0; j < 60; j++) {
      int x = D[u * 2][j];
      D[u][j] = x + D[u * 2 + 1][(x + j) % 60];
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int n;
  cin >> n;
  int suma = 1;
  while (suma < n) suma *= 2;
  suma *= 2;
  suma--;
  for (int i = 1; i <= n; i++) cin >> T[i];
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 60; j++) {
      D[i + suma / 2][j] = 1;
      if (j % T[i] == 0) D[i + suma / 2][j]++;
    }
  }
  for (int i = suma / 2; i >= 1; i--) {
    for (int j = 0; j < 60; j++) {
      int x = D[i * 2][j];
      D[i][j] = x + D[i * 2 + 1][(x + j) % 60];
    }
  }
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    char c;
    cin >> c;
    int x, y;
    cin >> x >> y;
    if (c == 'A')
      cout << getsum(x, y, 1, suma / 2 + 2, 1) << "\n";
    else
      update(x + suma / 2, y);
    last = 0;
  }
  return 0;
}
