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 N = 100500;
const int K = 4 * N;
const int T = 60;
pair<short, int> nxt[K][T];
pair<short, int> go[T][7];
int tl[K], tr[K], rev[N], a[N], n;
pair<short, int> ans;
void upd(int x) {
  for (int i = 0; i < 60; i++) {
    nxt[x][i] = make_pair(nxt[x + x + 2][nxt[x + x + 1][i].first].first,
                          nxt[x + x + 1][i].second +
                              nxt[x + x + 2][nxt[x + x + 1][i].first].second);
  }
}
void build(int x, int l, int r) {
  tl[x] = l;
  tr[x] = r;
  if (l < r) {
    int mid = (l + r) >> 1;
    build(x + x + 1, l, mid);
    build(x + x + 2, mid + 1, r);
    upd(x);
  } else {
    for (int i = 0; i < 60; i++) {
      nxt[x][i] = go[i][a[l]];
    }
    rev[l] = x;
  }
}
void solve(int x, int l, int r) {
  if ((tl[x] == l) && (tr[x] == r)) {
    ans = make_pair(nxt[x][ans.first].first,
                    ans.second + nxt[x][ans.first].second);
    return;
  }
  if (l <= tr[x + x + 1]) solve(x + x + 1, l, min(r, tr[x + x + 1]));
  if (r >= tl[x + x + 2]) solve(x + x + 2, max(l, tl[x + x + 2]), r);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < 60; i++) {
    for (int j = 2; j <= 6; j++) {
      if (i % j != 0) {
        go[i][j] = make_pair((i + 1) % 60, 1);
      } else {
        go[i][j] = make_pair((i + 2) % 60, 2);
      }
    }
  }
  build(0, 0, n - 1);
  int tt;
  scanf("%d\n", &tt);
  while (tt--) {
    char c;
    int x, y;
    scanf("%c %d %d\n", &c, &x, &y);
    if (c == 'A') {
      --x;
      --y;
      --y;
      ans.first = 0;
      ans.second = 0;
      solve(0, x, y);
      printf("%d\n", ans.second);
    } else {
      --x;
      a[x] = y;
      x = rev[x];
      for (int i = 0; i < 60; i++) nxt[x][i] = go[i][y];
      while (x > 0) {
        x = (x - 1) >> 1;
        upd(x);
      }
    }
  }
}
