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, p = 1, q;
int a[100005];
int seg[300005][60];
void fix(int at) {
  for (int j = 0; j < 60; j++) {
    int x = seg[2 * at][j];
    seg[at][j] = seg[2 * at + 1][(x + j) % 60] + x;
  }
}
void update(int at, int lo, int hi, int x) {
  if (hi < x || lo > x) return;
  if (hi == lo) {
    for (int j = 0; j < 60; j++) {
      seg[at][j] = 1 + (j % a[x] == 0);
    }
  } else {
    update(2 * at, lo, (lo + hi) / 2, x);
    update(2 * at + 1, (lo + hi + 1) / 2, hi, x);
    fix(at);
  }
}
int go(int at, int lo, int hi, int qlo, int qhi, int t) {
  if (hi < qlo || qhi < lo) return t;
  if (qlo <= lo && hi <= qhi) {
    return t + seg[at][t % 60];
  }
  int t1 = go(2 * at, lo, (lo + hi) / 2, qlo, qhi, t);
  return go(2 * at + 1, (lo + hi) / 2 + 1, hi, qlo, qhi, t1);
}
int main() {
  scanf("%d", &n);
  while (p < n) p *= 2;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    for (int j = 0; j < 60; j++) {
      seg[i + p][j] = 1 + (j % a[i] == 0);
    }
  }
  for (int i = p - 1; i >= 1; i--) {
    fix(i);
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    char s[10];
    int x, y;
    scanf("%s %d %d", s, &x, &y);
    if (s[0] == 'C') {
      x--;
      a[x] = y;
      update(1, 0, p - 1, x);
    } else {
      x--;
      y--;
      printf("%d\n", go(1, 0, p - 1, x, y - 1, 0));
    }
  }
}
