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;
template <typename T>
T read1() {
  T t = 0;
  char k;
  bool v = 0;
  do (k = getchar()) == '-' && (v = 1);
  while ('0' > k || k > '9');
  while ('0' <= k && k <= '9')
    t = (t << 3) + (t << 1) + (k ^ '0'), k = getchar();
  return v ? -t : t;
}
int v[400005][60], s, a[100005], ans[60];
void merge(int* x, int* u, int* v) {
  for (int i = 0; i < 60; ++i) x[i] = u[i] + v[(i + u[i]) % 60];
}
void up(int d) { merge(v[d], v[d << 1], v[d << 1 | 1]); }
void build(int l, int r, int d) {
  if (l == r) {
    a[l] = read1<long long>();
    for (int i = 0; i < 60; ++i)
      if (i - i / a[l] * a[l])
        v[d][i] = 1;
      else
        v[d][i] = 2;
  } else {
    int mid = l + r >> 1;
    build(l, mid, d << 1);
    build(mid + 1, r, d << 1 | 1);
    up(d);
  }
}
void add(int l, int r, int w, int d) {
  if (l == r) {
    a[l] = read1<long long>();
    for (int i = 0; i < 60; ++i)
      if (i - i / a[l] * a[l])
        v[d][i] = 1;
      else
        v[d][i] = 2;
  } else {
    int mid = l + r >> 1;
    if (w <= mid)
      add(l, mid, w, d << 1);
    else
      add(mid + 1, r, w, d << 1 | 1);
    up(d);
  }
}
void query(int l, int r, int o, int p, int d) {
  if (l == o && r == p)
    merge(ans, ans, v[d]);
  else {
    int mid = l + r >> 1;
    if (p <= mid)
      query(l, mid, o, p, d << 1);
    else if (mid < o)
      query(mid + 1, r, o, p, d << 1 | 1);
    else
      query(l, mid, o, mid, d << 1), query(mid + 1, r, mid + 1, p, d << 1 | 1);
  }
}
int main() {
  s = read1<long long>();
  build(1, s, 1);
  for (int T = read1<long long>(); T--;) {
    char k;
    cin >> k;
    if (k == 'A') {
      memset(ans, 0, sizeof(ans));
      int l = read1<long long>(), r = read1<long long>();
      if (l > r) swap(l, r);
      if (l == r) {
        puts("0");
        continue;
      }
      query(1, s, l, r - 1, 1);
      printf("%d\n", ans[0]);
    } else
      add(1, s, read1<long long>(), 1);
  }
  return 0;
}
