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 <typename TYPE>
inline TYPE sqr(const TYPE &a) {
  return a * a;
}
const long double EPS = 1e-9;
const long double PI = 2 * acos(0.0);
const int N = 100010;
const int S = 60;
int a[N];
int t[N * 5][S];
void make_node(int *node, int k) {
  for (int i = 0; i < S;) {
    node[i++] = 2;
    for (int j = 1; j < k; ++j) {
      node[i++] = 1;
    }
  }
}
void up(int *node, int *left, int *right) {
  for (int i = 0; i < S; ++i) {
    int tmp = left[i];
    node[i] = tmp + right[(i + tmp) % S];
  }
}
void build(int id, int lf, int rg) {
  if (lf == rg) {
    make_node(t[id], a[lf]);
    return;
  }
  int m = (lf + rg) / 2;
  int u = id << 1;
  int v = u + 1;
  build(u, lf, m);
  build(v, m + 1, rg);
  up(t[id], t[u], t[v]);
}
void update(int id, int lf, int rg, int pos, int val) {
  if (lf == rg) {
    make_node(t[id], val);
    return;
  }
  int m = (lf + rg) / 2;
  int u = id << 1;
  int v = u + 1;
  if (pos <= m)
    update(u, lf, m, pos, val);
  else
    update(v, m + 1, rg, pos, val);
  up(t[id], t[u], t[v]);
}
int get(int id, int lf, int rg, int fm, int to, int k) {
  if (lf == fm && rg == to) return t[id][k];
  int m = (lf + rg) / 2;
  int u = id << 1;
  int v = u + 1;
  int res = 0;
  if (fm <= m) res += get(u, lf, m, fm, min(m, to), k);
  if (to > m) res += get(v, m + 1, rg, max(m + 1, fm), to, (res + k) % S);
  return res;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  build(1, 0, n - 1);
  int m;
  scanf("%d\n", &m);
  for (int it = 0; it < m; ++it) {
    char c;
    int x, y;
    scanf("%c%d%d\n", &c, &x, &y);
    if (c == 'A') {
      --x, --y;
      printf("%d\n", get(1, 0, n - 1, x, y - 1, 0));
    } else {
      --x;
      update(1, 0, n - 1, x, y);
    }
  }
  return 0;
}
