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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 1e5 + 1000, K = 60;
const int INF = 1e9;
const int MAGIC = 50;
int a[N];
int b[N / MAGIC][K + 1];
int c[N];
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  for (int i = 0; i < N; i++) {
    c[i] = i / MAGIC;
    a[i] = INF;
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < N / MAGIC; i++) {
    for (int j = 0; j <= K; j++) {
      b[i][j] = j;
    }
  }
  auto go = [&](int &x, int y) {
    if (x % y == 0) x++;
    x++;
  };
  auto upd = [&](int pos) {
    for (int j = 0; j <= K; j++) {
      b[pos][j] = j;
    }
    for (int i = pos * MAGIC; i < (pos + 1) * MAGIC; i++) {
      for (int j = 0; j <= K; j++) {
        go(b[c[i]][j], a[i]);
      }
    }
  };
  for (int i = 0; i < N / MAGIC; i++) upd(i);
  int q;
  cin >> q;
  while (q--) {
    char type;
    cin >> type;
    if (type == 'A') {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      r--;
      int ans = 0;
      while (l <= r) {
        if (l % MAGIC == 0 && l + MAGIC - 1 <= r) {
          int rem = ans % K;
          ans += b[c[l]][rem] - rem;
          l += MAGIC;
        } else {
          go(ans, a[l]);
          l++;
        }
      }
      cout << ans << '\n';
    } else {
      int pos, x;
      cin >> pos >> x;
      pos--;
      a[pos] = x;
      upd(c[pos]);
    }
  }
}
