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 a[100004];
typedef struct node {
  int value[60];
  node() {}
  void init(int v) {
    for (int i = 0; i < (60); i++) value[i] = 1 + !(i % v);
  }
  void merge(const node& l, const node& r) {
    for (int i = 0; i < (60); i++)
      value[i] = l.value[i] + r.value[(i + l.value[i]) % 60];
  }
} node;
node st[4 * 100004];
void buildSt(int i, int l, int r) {
  if (l == r)
    st[i].init(a[l]);
  else {
    int m = ((l + r) >> 1);
    buildSt((i << 1), l, m);
    buildSt(((i << 1) + 1), m + 1, r);
    st[i].merge(st[(i << 1)], st[((i << 1) + 1)]);
  }
}
void update(int i, int s, int e, int pos) {
  if (s == e)
    st[i].init(a[s]);
  else {
    int m = ((s + e) >> 1);
    if (pos <= m)
      update((i << 1), s, m, pos);
    else
      update(((i << 1) + 1), m + 1, e, pos);
    st[i].merge(st[(i << 1)], st[((i << 1) + 1)]);
  }
}
node rangeQuery(int i, int s, int e, int l, int r) {
  if (l == s && e == r) return st[i];
  int m = ((s + e) >> 1);
  if (r <= m) return rangeQuery((i << 1), s, m, l, r);
  if (l > m) return rangeQuery(((i << 1) + 1), m + 1, e, l, r);
  node nd;
  nd.merge(rangeQuery((i << 1), s, m, l, min(m, r)),
           rangeQuery(((i << 1) + 1), m + 1, e, max(l, m + 1), r));
  return nd;
}
int main() {
  std::ios::sync_with_stdio(false);
  int n, q, x, y;
  char c;
  cin >> n;
  for (int i = 0; i < (n); i++) cin >> a[i];
  ;
  buildSt(1, 0, n - 1);
  cin >> q;
  for (int i = 0; i < (q); i++) {
    cin >> c >> x >> y;
    switch (c) {
      case 'A':
        cout << rangeQuery(1, 0, n - 1, x - 1, y - 2).value[0] << endl;
        break;
      case 'C':
        a[x - 1] = y;
        update(1, 0, n - 1, x - 1);
        break;
    }
  }
  return 0;
}
