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 = 100010;
int n, q, a[N];
char s[2];
struct State {
  int w[60];
  friend State operator+(const State &state_l, const State &state_r) {
    State ret;
    for (int i = 0; i < 60; ++i)
      ret.w[i] = state_l.w[i] + state_r.w[(i + state_l.w[i]) % 60];
    return ret;
  }
} state[N << 2];
void build(int c, int l, int r) {
  if (l == r) {
    for (int j = 0; j < 60; ++j) state[c].w[j] = (j % a[l] == 0 ? 2 : 1);
    return;
  }
  int mid = l + r >> 1;
  build(c << 1, l, mid), build(c << 1 | 1, mid + 1, r);
  state[c] = state[c << 1] + state[c << 1 | 1];
}
void change(int c, int l, int r, int pos) {
  if (l == r) {
    for (int j = 0; j < 60; ++j) state[c].w[j] = (j % a[l] == 0 ? 2 : 1);
    return;
  }
  int mid = l + r >> 1;
  if (pos <= mid)
    change(c << 1, l, mid, pos);
  else
    change(c << 1 | 1, mid + 1, r, pos);
  state[c] = state[c << 1] + state[c << 1 | 1];
}
State query(int c, int l, int r, int L, int R) {
  if (l == L && r == R) return state[c];
  int mid = l + r >> 1;
  if (R <= mid)
    return query(c << 1, l, mid, L, R);
  else if (L > mid)
    return query(c << 1 | 1, mid + 1, r, L, R);
  return query(c << 1, l, mid, L, mid) +
         query(c << 1 | 1, mid + 1, r, mid + 1, R);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  build(1, 1, n);
  scanf("%d", &q);
  for (int i = 0, x, y; i < q; ++i) {
    scanf("%s %d %d", s, &x, &y);
    if (*s == 'C') {
      a[x] = y;
      change(1, 1, n, x);
    } else {
      if (x == y)
        puts("0");
      else
        printf("%d\n", query(1, 1, n, x, y - 1).w[0]);
    }
  }
  return 0;
}
