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 N;
int A[100100];
struct node {
  int ans[60];
} seg[262144];
node combine(node a, node b) {
  node c;
  for (int i = 0; i < 60; i++) {
    c.ans[i] = a.ans[i];
    c.ans[i] += b.ans[(i + c.ans[i]) % 60];
  }
  return c;
}
void init(int idx, int begin, int end) {
  if (begin == end) {
    for (int i = 0; i < 60; i++) seg[idx].ans[i] = (i % A[begin] == 0) ? 2 : 1;
  } else {
    int mid = (begin + end) / 2;
    init(idx * 2, begin, mid);
    init(idx * 2 + 1, mid + 1, end);
    seg[idx] = combine(seg[idx * 2], seg[idx * 2 + 1]);
  }
}
void update(int idx, int begin, int end, int x, int v) {
  if (x < begin || end < x) return;
  if (begin == end) {
    A[begin] = v;
    for (int i = 0; i < 60; i++) seg[idx].ans[i] = (i % A[begin] == 0) ? 2 : 1;
  } else {
    int mid = (begin + end) / 2;
    update(idx * 2, begin, mid, x, v);
    update(idx * 2 + 1, mid + 1, end, x, v);
    seg[idx] = combine(seg[idx * 2], seg[idx * 2 + 1]);
  }
}
node query(int idx, int begin, int end, int i, int j) {
  if (j < begin || end < i) {
    node dummy;
    dummy.ans[0] = -1;
    return dummy;
  }
  if (i <= begin && end <= j) return seg[idx];
  int mid = (begin + end) / 2;
  node a = query(idx * 2, begin, mid, i, j);
  node b = query(idx * 2 + 1, mid + 1, end, i, j);
  if (a.ans[0] == -1) return b;
  if (b.ans[0] == -1) return a;
  return combine(a, b);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", A + i);
  int Q;
  scanf("%d", &Q);
  int a, b;
  char c;
  init(1, 1, N);
  for (int i = 0; i < Q; i++) {
    scanf(" %c%d%d", &c, &a, &b);
    if (c == 'C')
      update(1, 1, N, a, b);
    else
      printf("%d\n", query(1, 1, N, a, b - 1).ans[0]);
  }
  return 0;
}
