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 q, n, m, a[100001];
vector<int> t[600001];
vector<int> merge(vector<int> v1, vector<int> v2) {
  vector<int> p;
  if (!v1.size()) v1.resize(60);
  if (!v2.size()) v2.resize(60);
  for (int k = 0; k < 60; ++k) {
    int t1 = v1[k];
    t1 += v2[(k + t1) % 60];
    p.push_back(t1);
  }
  return p;
}
void upd(int v, int l, int r, int pos, int tt) {
  if (l == r) {
    t[v].resize(60);
    for (int k = 0; k < 60; k++) {
      t[v][k] = 1 + ((k % tt) == 0);
    }
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    upd(v + v, l, mid, pos, tt);
  else
    upd(v + v + 1, mid + 1, r, pos, tt);
  for (int k = 0; k < 60; ++k) {
    int t1 = t[2 * v][k];
    t1 += t[2 * v + 1][(k + t1) % 60];
    t[v][k] = t1;
  }
  return;
}
int tim = 0;
int get(int v, int l, int r, int l1, int r1) {
  if (l > r1 || r < l1) return 0;
  if (l == l1 && r == r1) {
    if (!t[v].size()) t[v].resize(60);
    int yy = tim;
    tim = tim + t[v][yy % 60];
    return t[v][yy % 60];
  }
  int mid = (l + r) >> 1;
  int A, B;
  A = get(v + v, l, mid, l1, min(r1, mid));
  B = get(v + v + 1, mid + 1, r, max(mid + 1, l1), r1);
  return A + B;
}
void build(int v, int l, int r) {
  if (l == r) {
    t[v].resize(60, 0);
    for (int k = 0; k < 60; ++k) {
      t[v][k] = 1;
      if ((k % a[l]) == 0) ++t[v][k];
    }
    return;
  }
  int mid = (l + r) / 2;
  build(2 * v, l, mid);
  build(2 * v + 1, mid + 1, r);
  t[v] = merge(t[2 * v], t[2 * v + 1]);
  return;
}
int main() {
  scanf("%d", &n);
  ;
  for (int k = 1; k <= n; ++k) {
    scanf("%d", &a[k]);
    ;
  }
  build(1, 1, n);
  scanf("%d", &q);
  ;
  while (q--) {
    char op;
    int x, y;
    scanf("%c", &op);
    scanf("%c", &op);
    scanf("%d", &x);
    ;
    scanf("%d", &y);
    ;
    if (op == 'C') {
      upd(1, 1, n, x, y);
      continue;
    }
    tim = 0;
    int ans = get(1, 1, n, x, y - 1);
    printf("%d\n", tim);
    ;
  }
  return 0;
}
