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;
struct node {
  int left, right;
  int time[60];
  node(int left = 0, int right = 0) : left(left), right(right) {}
} arb[4 * 100000 + 5];
inline void unite(node &a, node b, node &c) {
  for (int i = 0; i < 60; i++)
    a.time[i] = b.time[i] + c.time[(i + b.time[i]) % 60];
}
int a[100000 + 5];
void build(int left, int right, int pos) {
  arb[pos].left = left, arb[pos].right = right;
  if (left == right) {
    for (int i = 0; i < 60; i++)
      if (i % a[left])
        arb[pos].time[i] = 1;
      else
        arb[pos].time[i] = 2;
    return;
  }
  int mijl = (left + right) >> 1;
  build(left, mijl, pos << 1);
  build(mijl + 1, right, (pos << 1) + 1);
  unite(arb[pos], arb[pos << 1], arb[(pos << 1) + 1]);
}
void update(int up_pos, int val, int pos) {
  if (arb[pos].left == arb[pos].right) {
    for (int i = 0; i < 60; i++)
      if (i % val)
        arb[pos].time[i] = 1;
      else
        arb[pos].time[i] = 2;
    return;
  }
  int mijl = (arb[pos].left + arb[pos].right) >> 1;
  if (up_pos <= mijl)
    update(up_pos, val, pos << 1);
  else
    update(up_pos, val, (pos << 1) + 1);
  unite(arb[pos], arb[pos << 1], arb[(pos << 1) + 1]);
}
int ans;
void query(int x, int y, int pos) {
  if (arb[pos].left == x && arb[pos].right == y) {
    ans += arb[pos].time[ans % 60];
    return;
  }
  int mijl = (arb[pos].left + arb[pos].right) >> 1;
  if (y <= mijl)
    query(x, y, pos << 1);
  else if (x > mijl)
    query(x, y, (pos << 1) + 1);
  else {
    query(x, mijl, pos << 1);
    query(mijl + 1, y, (pos << 1) + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, n, 1);
  int q = 0;
  cin >> q;
  char type;
  int x, y;
  while (q--) {
    cin >> type >> x >> y;
    if (type == 'A') {
      y--;
      ans = 0;
      query(x, y, 1);
      cout << ans << '\n';
    } else
      update(x, y, 1);
  }
  return 0;
}
