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;
static int m1[3 * (1 << 17)][60], m2[2 * (1 << 17)];
inline void up(int n) {
  if (n >= (1 << 17)) return;
  for (int i = 59; i >= 0; --i) {
    m1[n][i] = m1[n * 2][i] + m1[n * 2 + 1][(i + m1[n * 2][i]) % 60];
  }
}
void in(int v, int d) {
  d += (1 << 17);
  for (int i = 0; i < 60; ++i)
    if (i % v == 0)
      m1[d][i] = 2;
    else
      m1[d][i] = 1;
  for (int i = d; i > 0; i /= 2) up(i);
}
int gt(int cu, int l, int r, int lx, int rx, int mo) {
  if (lx >= r || l >= rx) return 0;
  if (lx <= l && rx >= r) {
    return m1[cu][mo];
  }
  int a = gt(cu * 2, l, (l + r) / 2, lx, rx, mo);
  return a + gt(cu * 2 + 1, (l + r) / 2, r, lx, rx, (a + mo) % 60);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, a, b, m;
  char ch;
  scanf("%i", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%i", &m2[i]);
    in(m2[i], i + 1);
  }
  scanf("%i\n", &m);
  for (int i = 0; i < m; ++i) {
    scanf("%c%i%i\n", &ch, &a, &b);
    if (ch == 'A')
      printf("%i\n", gt(1, 0, (1 << 17), a, b, 0));
    else
      in(b, a);
  }
  return 0;
}
