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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
const int N = 1 << 17;
int n;
int m;
pair<int, int> tree[2 * N + 10][60];
pair<int, int> y[60];
pair<int, int> cur[60];
int p[100000];
int calc(int x, int y) {
  int p = ::p[y];
  for (int k = 0; k < 60; k++)
    if (k % p == 0)
      tree[x][k] = make_pair(2, (k + 2) % 60);
    else
      tree[x][k] = make_pair(1, (k + 1) % 60);
  return 0;
}
int merge(pair<int, int> *x, pair<int, int> *a, pair<int, int> *b) {
  for (int k = 0; k < 60; k++)
    y[k] = make_pair(a[k].first + b[a[k].second].first, b[a[k].second].second);
  for (int k = 0; k < 60; k++) x[k] = y[k];
  return 0;
}
int build(int x, int l, int r) {
  if (l == r) {
    calc(x, l);
    return 0;
  }
  int s = (l + r) / 2;
  build(x * 2 + 1, l, s);
  build(x * 2 + 2, s + 1, r);
  merge(tree[x], tree[x * 2 + 1], tree[x * 2 + 2]);
  return 0;
}
int upd(int x, int l, int r, int y, int z) {
  if (r < y || l > y) return 0;
  if (l == r) {
    p[y] = z;
    calc(x, y);
    return 0;
  }
  int s = (l + r) / 2;
  upd(x * 2 + 1, l, s, y, z);
  upd(x * 2 + 2, s + 1, r, y, z);
  merge(tree[x], tree[x * 2 + 1], tree[x * 2 + 2]);
  return 0;
}
int get(int x, int l, int r, int lc, int rc) {
  if (r < lc || l > rc) return 0;
  if (l >= lc && r <= rc) {
    merge(cur, cur, tree[x]);
    return 0;
  }
  int s = (l + r) / 2;
  get(x * 2 + 1, l, s, lc, rc);
  get(x * 2 + 2, s + 1, r, lc, rc);
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &p[i]);
  build(0, 0, n - 1);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    char h[5];
    int a, b;
    scanf(" %s%d%d", h, &a, &b);
    if (h[0] == 'A') {
      for (int j = 0; j < 60; j++) cur[j] = make_pair(0, j);
      get(0, 0, n - 1, a - 1, b - 2);
      printf("%d\n", cur[0].first);
    } else {
      upd(0, 0, n - 1, a - 1, b);
    }
  }
  return 0;
}
