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 N = 100005;
const int K = 65;
const int inf = 1000 * 1000 * 1000;
const int mod = 1000 * 1000 * 1000 + 7;
int n, a[N], q;
int t[4 * N][K];
void build(int v, int s, int e) {
  if (s == e) {
    for (int i = 0; i < 60; i++) {
      if (i % a[s] == 0)
        t[v][i] = 2;
      else
        t[v][i] = 1;
    }
    return;
  }
  int m = (s + e) / 2;
  build(2 * v, s, m);
  build(2 * v + 1, m + 1, e);
  for (int i = 0; i < 60; i++) {
    int x = (i + t[2 * v][i]) % 60;
    t[v][i] = t[2 * v][i] + t[2 * v + 1][x];
  }
}
void update(int v, int s, int e, int pos) {
  if (s == e) {
    for (int i = 0; i < 60; i++) {
      if (i % a[s] == 0)
        t[v][i] = 2;
      else
        t[v][i] = 1;
    }
    return;
  }
  int m = (s + e) / 2;
  if (pos <= m)
    update(2 * v, s, m, pos);
  else
    update(2 * v + 1, m + 1, e, pos);
  for (int i = 0; i < 60; i++) {
    int x = (i + t[2 * v][i]) % 60;
    t[v][i] = t[2 * v][i] + t[2 * v + 1][x];
  }
}
int get(int v, int s, int e, int l, int r, int st) {
  if (l > r) return -1;
  if (s == l && e == r) return t[v][st];
  int m = (s + e) / 2;
  int answ = 0;
  int T = get(2 * v, s, m, l, min(r, m), st);
  if (T == -1)
    T = st;
  else {
    answ += T;
    T += st;
  }
  int P = get(2 * v + 1, m + 1, e, max(m + 1, l), r, T % 60);
  if (P != -1) answ += P;
  return answ;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[n + 1] = 1;
  build(1, 1, n + 1);
  scanf("%d", &q);
  while (q--) {
    char x;
    int aa, b;
    cin >> x;
    scanf("%d%d", &aa, &b);
    if (x == 'A') {
      int T = get(1, 1, n + 1, aa, b - 1, 0);
      printf("%d\n", T);
    } else {
      a[aa] = b;
      update(1, 1, n + 1, aa);
    }
  }
  return 0;
}
