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 = 100001;
const int MD = 60;
int a[MaxN];
int rmq[4 * MaxN][MD];
void ver1(int x, int* ans) {
  for (int i = 0; i < MD; i++) {
    if (i % x == 0) {
      ans[i] = 2;
    } else {
      ans[i] = 1;
    }
  }
}
void merge(int* n1, int* n2, int* ans) {
  for (int i = 0; i < MD; i++) {
    ans[i] = n1[i] + n2[(i + n1[i]) % MD];
  }
}
int lv, rv;
void get_rmq(int v, int l, int r, int* ans) {
  if ((lv <= l) && (r <= rv)) {
    for (int i = 0; i < MD; i++) {
      ans[i] = rmq[v][i];
    }
  } else {
    int md = (l + r) / 2;
    if (rv <= md) {
      get_rmq(v * 2 + 1, l, md, ans);
    } else if (lv > md) {
      get_rmq(v * 2 + 2, md + 1, r, ans);
    } else {
      int a1[MD], a2[MD];
      get_rmq(v * 2 + 1, l, md, a1);
      get_rmq(v * 2 + 2, md + 1, r, a2);
      merge(a1, a2, ans);
    }
  }
}
void modify_rmq(int v, int l, int r) {
  if (l == r) {
    ver1(rv, rmq[v]);
  } else {
    if (lv <= (l + r) / 2) {
      modify_rmq(v * 2 + 1, l, (l + r) / 2);
    } else {
      modify_rmq(v * 2 + 2, (l + r) / 2 + 1, r);
    }
    merge(rmq[v * 2 + 1], rmq[v * 2 + 2], rmq[v]);
  }
}
void create_rmq(int v, int l, int r) {
  if (l == r) {
    ver1(a[l], rmq[v]);
  } else {
    create_rmq(v * 2 + 1, l, (l + r) / 2);
    create_rmq(v * 2 + 2, (l + r) / 2 + 1, r);
    merge(rmq[v * 2 + 1], rmq[v * 2 + 2], rmq[v]);
  }
}
int n, q;
int av[MaxN];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  create_rmq(0, 1, n);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    char c, d;
    scanf("%c %c %d %d", &d, &c, &lv, &rv);
    if (c == 'C') {
      modify_rmq(0, 1, n);
    } else {
      rv--;
      get_rmq(0, 1, n, av);
      printf("%d\n", av[0]);
    }
  }
  return 0;
}
