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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
int a[100010];
struct node {
  int v[60];
} t[100010 << 2];
void up(int p) {
  for (int i = 0; i < 60; i++) {
    int first = t[p << 1].v[i];
    int second = (i + first) % 60;
    t[p].v[i] = first + t[p << 1 | 1].v[second];
  }
}
void build(int p, int l, int r) {
  for (int i = 0; i < 60; i++) t[p].v[i] = 0;
  if (l == r) {
    for (int i = 0; i < 60; i++) {
      int j = i, k = 0;
      while (j % a[l] == 0) k++, j++;
      t[p].v[i] = k + 1;
    }
    return;
  }
  int m = (l + r) >> 1;
  build(p << 1, l, m);
  build(p << 1 | 1, m + 1, r);
  up(p);
}
void upd(int p, int l, int r, int first, int second) {
  if (l == r) {
    for (int i = 0; i < 60; i++) {
      int j = i, k = 0;
      while (j % second == 0) k++, j++;
      t[p].v[i] = k + 1;
    }
    return;
  }
  int m = (l + r) >> 1;
  if (first <= m)
    upd(p << 1, l, m, first, second);
  else
    upd(p << 1 | 1, m + 1, r, first, second);
  up(p);
}
int query(int p, int l, int r, int first, int second, int k) {
  if (l >= first && r <= second) return t[p].v[k];
  int m = (l + r) >> 1, ans = 0;
  if (second <= m) return query(p << 1, l, m, first, second, k);
  if (first > m) return query(p << 1 | 1, m + 1, r, first, second, k);
  int res = query(p << 1, l, m, first, m, k);
  ans = res;
  res = (k + res) % 60;
  ans += query(p << 1 | 1, m + 1, r, m + 1, second, res);
  return ans;
}
int main() {
  int i, j, k, ca = 0, m, n;
  while (~scanf("%d", &n)) {
    for (i = 0; i < n; i++) scanf("%d", &a[i]);
    build(1, 0, n - 1);
    scanf("%d", &m);
    while (m--) {
      char s[2];
      int first, second;
      scanf("%s%d%d", s, &first, &second);
      if (s[0] == 'C') {
        first--;
        upd(1, 0, n - 1, first, second);
      } else {
        first--, second -= 2;
        int ans = query(1, 0, n - 1, first, second, 0);
        printf("%d\n", ans);
      }
    }
  }
}
