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:1000000")
using namespace std;
int c, v, c1, v1, *a[262144], *s[262144];
void l(int w, int e, int t) {
  w <<= 1;
  if ((c <= e) && (e + t - 1 <= v)) {
    c1 += a[w][v1];
    v1 = s[w][v1];
  } else if (min(v, e + t - 1) - max(c, e) >= 0)
    l(w, e, t >> 1);
  if ((c <= e + t) && (e + t + t - 1 <= v)) {
    c1 += a[w + 1][v1];
    v1 = s[w + 1][v1];
  } else if (min(v, e + t + t - 1) - max(c, e + t) >= 0)
    l(w + 1, e + t, t >> 1);
}
int main() {
  int q, w, e, r, t, o;
  for (w = 0; w < 262144; w++) {
    a[w] = new int[60];
    s[w] = new int[60];
  }
  cin >> q;
  for (w = 0; w < q; w++) {
    cin >> e;
    for (r = 0; r < 60; r++) {
      c = 1;
      if (r % e == 0) c++;
      a[w + 131072][r] = c;
      s[w + 131072][r] = (c + r) % 60;
    }
  }
  for (; w < 131072; w++)
    for (r = 0; r < 60; r++) a[w + 131072][r] = s[w + 131072][r] = 1;
  for (w = 131071; w; w--)
    for (e = 0; e < 60; e++) {
      c = s[w << 1][e];
      a[w][e] = a[w << 1][e] + a[(w << 1) + 1][c];
      s[w][e] = s[(w << 1) + 1][c];
    }
  char d[2];
  for (cin >> o; o; o--) {
    scanf("%s%d%d", &d, &c, &v);
    if (d[0] == 'A') {
      c1 = v1 = 0;
      c--;
      v -= 2;
      l(1, 0, 65536);
      printf("%d\n", c1);
    } else {
      c += 131071;
      for (e = 0; e < 60; e++) {
        r = 1;
        if (e % v == 0) r++;
        a[c][e] = r;
        s[c][e] = (r + e) % 60;
      }
      for (c >>= 1; c; c >>= 1)
        for (e = 0; e < 60; e++) {
          r = s[c << 1][e];
          a[c][e] = a[c << 1][e] + a[(c << 1) + 1][r];
          s[c][e] = s[(c << 1) + 1][r];
        }
    }
  }
  return 0;
}
