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;
const int Maxd = 60;
const int Maxn = 100005;
const int Maxm = 524288;
int n;
int a[Maxn];
int st[Maxm][Maxd];
int q;
void Union(int v) {
  for (int j = 0; j < Maxd; j++) {
    int t1 = st[2 * v][j];
    int t2 = st[2 * v + 1][(j + t1) % Maxd];
    st[v][j] = t1 + t2;
  }
}
void Create(int v, int l, int r) {
  if (l == r)
    for (int j = 0; j < Maxd; j++) st[v][j] = j % a[l] == 0 ? 2 : 1;
  else {
    int m = l + r >> 1;
    Create(2 * v, l, m);
    Create(2 * v + 1, m + 1, r);
    Union(v);
  }
}
void Update(int v, int l, int r, int x) {
  if (l == r)
    for (int j = 0; j < Maxd; j++) st[v][j] = j % a[l] == 0 ? 2 : 1;
  else {
    int m = l + r >> 1;
    if (x <= m)
      Update(2 * v, l, m, x);
    else
      Update(2 * v + 1, m + 1, r, x);
    Union(v);
  }
}
int Calc(int v, int l, int r, int a, int b, int cur = 0) {
  if (l == a && r == b)
    return st[v][cur];
  else {
    int m = l + r >> 1;
    if (b <= m) return Calc(2 * v, l, m, a, b, cur);
    if (m + 1 <= a) return Calc(2 * v + 1, m + 1, r, a, b, cur);
    int t1 = Calc(2 * v, l, m, a, m, cur);
    cur = (cur + t1) % Maxd;
    int t2 = Calc(2 * v + 1, m + 1, r, m + 1, b, cur);
    return t1 + t2;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  Create(1, 1, n);
  scanf("%d", &q);
  char ch;
  int x, y;
  while (q--) {
    scanf(" %c %d %d", &ch, &x, &y);
    if (ch == 'C') {
      a[x] = y;
      Update(1, 1, n, x);
    } else {
      y--;
      printf("%d\n", Calc(1, 1, n, x, y));
    }
  }
  return 0;
}
