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;
int H[60][100100 << 2], sol, v[100100], n, m, x, y, l, r;
char ch;
void huehue(int x) {
  int lef = (x << 1);
  int rig = (x << 1) + 1;
  for (int j = 0; j <= 59; ++j) {
    int costu = H[j][lef];
    int cost = H[j][lef];
    cost += j;
    cost %= 60;
    costu += H[cost][rig];
    H[j][x] = costu;
  }
}
void go(int st, int dr, int po) {
  if (st == dr) {
    for (int j = 0; j <= 59; ++j)
      if (j % v[st] == 0)
        H[j][po] = 2;
      else
        H[j][po] = 1;
    return;
  }
  int mij = (st + dr) >> 1;
  go(st, mij, po << 1);
  go(mij + 1, dr, (po << 1) + 1);
  huehue(po);
}
void upd(int st, int dr, int po, int poz, int x) {
  if (st == dr) {
    for (int j = 0; j <= 59; ++j)
      if (j % x == 0)
        H[j][po] = 2;
      else
        H[j][po] = 1;
    return;
  }
  int mij = (st + dr) >> 1;
  if (poz <= mij)
    upd(st, mij, po << 1, poz, x);
  else
    upd(mij + 1, dr, (po << 1) + 1, poz, x);
  huehue(po);
}
void query(int st, int dr, int po, int l, int r) {
  if (st >= l && dr <= r) {
    sol += H[sol % 60][po];
    return;
  }
  int mij = (st + dr) >> 1;
  if (l <= mij) query(st, mij, po << 1, l, r);
  if (r > mij) query(mij + 1, dr, (po << 1) + 1, l, r);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> v[i];
  go(1, n, 1);
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> ch;
    if (ch == 'A') {
      sol = 0;
      cin >> l >> r;
      --r;
      query(1, n, 1, l, r);
      cout << sol << "\n";
    } else {
      cin >> x >> y;
      upd(1, n, 1, x, y);
    }
  }
  return 0;
}
