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 MAXV = 60;
const int N = 100005;
int tree[N * 4][MAXV + 5], leaf[N];
int a[N];
void combine(int nd) {
  int cc = nd << 1;
  for (int i = 0; i < MAXV; i++)
    tree[nd][i] = tree[cc][i] + tree[cc ^ 1][(tree[cc][i] + i) % MAXV];
}
void build(int nd, int le, int ri) {
  if (le == ri) {
    for (int i = 0; i < MAXV; i++)
      if (i % a[le] == 0)
        tree[nd][i] = 2;
      else
        tree[nd][i] = 1;
    leaf[le] = nd;
    return;
  }
  int mi = (le + ri) >> 1;
  int cc = nd << 1;
  build(cc, le, mi);
  build(cc ^ 1, mi + 1, ri);
  combine(nd);
}
void update(int id, int val) {
  if (a[id] == val) return;
  a[id] = val;
  int nd = leaf[id];
  for (int i = 0; i < MAXV; i++)
    if (i % val == 0)
      tree[nd][i] = 2;
    else
      tree[nd][i] = 1;
  nd >>= 1;
  while (nd > 0) {
    combine(nd);
    nd >>= 1;
  }
}
pair<int, int> query(int nd, int le, int ri, int tr, int ph, int val) {
  if (le > ph || ri < tr) return {0, 0};
  if (tr <= le && ri <= ph) {
    return {tree[nd][val], (val + tree[nd][val]) % MAXV};
  }
  int mi = (le + ri) >> 1;
  int cc = nd << 1;
  auto p = query(cc, le, mi, tr, ph, val);
  auto p2 = query(cc ^ 1, mi + 1, ri, tr, ph, p.second);
  return {p.first + p2.first, p2.second};
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  build(1, 1, n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y;
    char code[10];
    scanf("%s", code);
    scanf("%d%d", &x, &y);
    if (code[0] == 'A') {
      printf("%d\n", query(1, 1, n, x, y - 1, 0));
    } else {
      update(x, y);
    }
  }
  return 0;
}
