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;
const int MAXN = 100000 + 10;
const int INF = 1000000000;
int a[MAXN];
int T[100][4 * MAXN];
int n, q;
int ans;
int calc(int k, int x) {
  int ret = 1;
  while (k % x == 0) k++, ret++;
  return ret;
}
void init(int x, int i, int j) {
  if (i == j) {
    for (int k = 0; k < 60; k++) T[k][x] = calc(k, a[i]);
    return;
  }
  int mid = (i + j) / 2;
  init(x + x, i, mid);
  init(x + x + 1, mid + 1, j);
  for (int k = 0; k < 60; k++) {
    T[k][x] = T[k][x + x] + T[(k + T[k][x + x]) % 60][x + x + 1];
  }
}
void update(int x, int i, int j, int u, int v) {
  if ((j < u) || (i > u)) return;
  if (i == j) {
    a[u] = v;
    for (int k = 0; k < 60; k++) T[k][x] = calc(k, a[i]);
    return;
  }
  int mid = (i + j) / 2;
  update(x + x, i, mid, u, v);
  update(x + x + 1, mid + 1, j, u, v);
  for (int k = 0; k < 60; k++) {
    T[k][x] = T[k][x + x] + T[(k + T[k][x + x]) % 60][x + x + 1];
  }
}
void get(int x, int i, int j, int u, int v) {
  if ((j < u) || (i > v)) return;
  if ((i >= u) && (j <= v)) {
    ans += T[ans % 60][x];
    return;
  }
  int mid = (i + j) / 2;
  get(x + x, i, mid, u, v);
  get(x + x + 1, mid + 1, j, u, v);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  init(1, 1, n);
  cin >> q;
  for (int i = 1; i <= q; i++) {
    char c;
    int x, y;
    cin >> c >> x >> y;
    if (c == 'C')
      update(1, 1, n, x, y);
    else {
      ans = 0;
      get(1, 1, n, x, y - 1);
      cout << ans << endl;
    }
  }
}
