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 tree[60][400005] = {0};
struct STree {
  int n;
  void init(int _n) { n = _n; }
  void add(int idx, int x, int k, int l, int r) {
    if (idx < l || r <= idx) return;
    if (idx == l && l + 1 == r) {
      for (int i = 0; i < 60; ++i) tree[i][k] = (i % x == 0 ? 2 : 1);
    } else {
      add(idx, x, k * 2 + 1, l, (l + r) / 2);
      add(idx, x, k * 2 + 2, (l + r) / 2, r);
      for (int i = 0; i < 60; ++i) {
        tree[i][k] =
            tree[i][k * 2 + 1] + tree[(i + tree[i][k * 2 + 1]) % 60][k * 2 + 2];
      }
    }
  }
  void add(int idx, int x) { add(idx, x, 0, 0, n); }
  int find(int t, int a, int b, int k, int l, int r) {
    if (b <= l || r <= a)
      return 0;
    else if (a <= l && r <= b)
      return tree[t][k];
    else {
      int ret = 0, tmp;
      tmp = find(t, a, b, k * 2 + 1, l, (l + r) / 2), ret += tmp;
      t = (t + tmp) % 60;
      tmp = find(t, a, b, k * 2 + 2, (l + r) / 2, r), ret += tmp;
      return ret;
    }
  }
  int find(int a, int b) { return find(0, a, b, 0, 0, n); }
} T;
int n, a[100005], q;
int main() {
  scanf("%d", &n), T.init(n + 5);
  for (int i = 0; i < n; ++i) scanf("%d", a + i), T.add(i, a[i]);
  scanf("%d", &q);
  while (q--) {
    char tmp[3];
    int x, y;
    scanf("%s%d%d", tmp, &x, &y);
    if (tmp[0] == 'A') {
      x--, y--;
      printf("%d\n", T.find(x, y));
    } else {
      x--, T.add(x, y);
    }
  }
}
