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 N, Q;
int type[100010];
struct node {
  node *L, *R;
  int lo, hi;
  int cost[60];
  node(int _lo = 0, int _hi = 0, node* _L = NULL, node* _R = NULL) {
    lo = _lo, hi = _hi;
    L = _L, R = _R;
  }
  void set(int t) {
    fill(cost, cost + 60, 1);
    for (int i = 0; i < 60; i += t) cost[i] = 2;
  }
  void merge() {
    for (int i = 0; i < 60; ++i)
      cost[i] = L->cost[i] + R->cost[(i + L->cost[i]) % 60];
  }
};
node* build(int fr, int to) {
  if (fr == to) {
    node* ret = new node(fr, to);
    ret->set(type[fr]);
    return ret;
  }
  node* ret = new node(fr, to, build(fr, (fr + to) >> 1),
                       build(((fr + to) >> 1) + 1, to));
  ret->merge();
  return ret;
}
int get(node* vr, int f, int t, int go = 0) {
  if (!vr || t < vr->lo || vr->hi < f) return go;
  if (f <= vr->lo && vr->hi <= t) return go + vr->cost[go % 60];
  return get(vr->R, f, t, get(vr->L, f, t, go));
}
void update(node* vr, int pos, int neo) {
  if (!vr || pos < vr->lo || vr->hi < pos) return;
  if (vr->lo == vr->hi) return vr->set(neo);
  update(vr->L, pos, neo);
  update(vr->R, pos, neo);
  vr->merge();
}
int main() {
  char buf[5];
  int x, y;
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) scanf("%d", &type[i]);
  node* root = build(0, N - 1);
  scanf("%d", &Q);
  for (int i = 0; i < Q; ++i) {
    scanf("%s%d%d", &buf, &x, &y);
    if (buf[0] == 'A')
      printf("%d\n", get(root, x - 1, y - 2, 0));
    else
      update(root, x - 1, y);
  }
  return 0;
}
