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 = 1e5 + 7;
const int M = 60;
const int inf = 1e9 + 7;
const long long base = 1e18;
const double pi = acos(-1);
const double ep = 1e-18;
int seg[M][N * 3];
int n;
void update(int i, int l, int r, int u, int x) {
  if (l == u && r == u) {
    for (int j = 0; j < M; j++) {
      if (j % x == 0)
        seg[j][i] = 2;
      else
        seg[j][i] = 1;
    }
    return;
  }
  int g = (l + r) >> 1, next = i << 1;
  if (u <= g)
    update(next, l, g, u, x);
  else
    update(next + 1, g + 1, r, u, x);
  for (int j = 0; j < M; j++) {
    int tmp = seg[j][next];
    seg[j][i] = tmp + seg[(j + tmp) % 60][next + 1];
  }
  return;
}
int get(int i, int l, int r, int u, int v, int first) {
  if (l == u && r == v) {
    return seg[first % 60][i];
  }
  int g = (l + r) >> 1, next = i << 1;
  int res = 0;
  if (u <= g) res += get(next, l, g, u, min(g, v), first);
  if (g < v) {
    if (u <= g) first = (first + res) % M;
    res += get(next + 1, g + 1, r, max(u, g + 1), v, first);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) {
    int u;
    scanf("%d", &u);
    update(1, 1, n, i, u);
  }
  int query;
  cin >> query;
  while (query-- > 0) {
    char c;
    int u, v;
    cin >> c >> u >> v;
    if (c == 'C') {
      update(1, 1, n, u, v);
    } else {
      printf("%d\n", get(1, 1, n, u, v - 1, 0));
    }
  }
  return 0;
}
