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 N = 100005, M = 60;
struct Node {
  int tm[M];
  Node(int x = 2) {
    for (int i = 0; i < (M); ++i) {
      int cur = i;
      while (cur % x == 0) ++cur;
      tm[i] = cur - i + 1;
    }
  }
  Node operator+(const Node& y) const {
    Node res;
    for (int i = 0; i < (M); ++i) res.tm[i] = tm[i] + y.tm[(i + tm[i]) % M];
    return res;
  }
} seg[N << 2], res;
int n, m;
void modify(int c, int l, int r, int p, int x) {
  if (l == r)
    seg[c] = Node(x);
  else {
    int m = (l + r) >> 1;
    if (p <= m)
      modify(c << 1, l, m, p, x);
    else
      modify(c << 1 | 1, m + 1, r, p, x);
    seg[c] = seg[c << 1] + seg[c << 1 | 1];
  }
}
void query(int c, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr)
    res = res + seg[c];
  else {
    int m = (l + r) >> 1;
    if (ql <= m) query(c << 1, l, m, ql, qr);
    if (qr > m) query(c << 1 | 1, m + 1, r, ql, qr);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < (n + 1); ++i) scanf("%d", &m), modify(1, 1, n, i, m);
  scanf("%d", &m);
  for (int i = 0; i < (m); ++i) {
    char s[10];
    int l, r;
    scanf("%s%d%d", s, &l, &r);
    if (s[0] == 'C')
      modify(1, 1, n, l, r);
    else {
      for (int i = 0; i < (M); ++i) res.tm[i] = 0;
      query(1, 1, n, l, r - 1);
      printf("%d\n", res.tm[0]);
    }
  }
  return 0;
}
