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 MAXN = 100005;
int a[MAXN];
int to[MAXN << 2][60];
int sum[MAXN << 2][60];
int get(int x) {
  if (x >= 60) x -= 60;
  return x;
}
void pushup(int o) {
  for (int i = 0; i < 60; i++) {
    to[o][i] = to[o << 1 | 1][to[o << 1][i]];
    sum[o][i] = sum[o << 1][i] + sum[o << 1 | 1][to[o << 1][i]];
  }
}
void build(int l, int r, int o) {
  if (l == r) {
    for (int i = 0; i < 60; i++) {
      if (i % a[l] == 0)
        to[o][i] = get(i + 2), sum[o][i] = 2;
      else
        to[o][i] = get(i + 1), sum[o][i] = 1;
    }
    return;
  }
  int m = l + r >> 1;
  build(l, m, o << 1);
  build(m + 1, r, o << 1 | 1);
  pushup(o);
}
void update(int l, int r, int o, int p, int v) {
  if (l == r) {
    a[l] = v;
    for (int i = 0; i < 60; i++) {
      if (i % a[l] == 0)
        to[o][i] = get(i + 2), sum[o][i] = 2;
      else
        to[o][i] = get(i + 1), sum[o][i] = 1;
    }
    return;
  }
  int m = l + r >> 1;
  if (p <= m)
    update(l, m, o << 1, p, v);
  else
    update(m + 1, r, o << 1 | 1, p, v);
  pushup(o);
}
int go, val;
void query(int l, int r, int o, int L, int R) {
  if (L <= l && r <= R) {
    val += sum[o][go];
    go = to[o][go];
    return;
  }
  int m = l + r >> 1;
  if (L <= m) query(l, m, o << 1, L, R);
  if (m < R) query(m + 1, r, o << 1 | 1, L, R);
}
int main() {
  int n, q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  build(1, n, 1);
  scanf("%d", &q);
  while (q--) {
    char s[2];
    int l, r;
    scanf("%s%d%d", s, &l, &r);
    if (s[0] == 'A') {
      go = 0;
      val = 0;
      query(1, n, 1, l, r - 1);
      printf("%d\n", val);
    } else
      update(1, n, 1, l, r);
  }
  return 0;
}
