#include <bits/stdc++.h>
using namespace std;
int encode(int t) { return (t % 4) * 30 + (t % 5) * 6 + (t % 6); }
int decode[120];
void gen() {
  memset(decode, -1, sizeof decode);
  for (int i = 0; i < 120; ++i) {
    int enc = encode(i);
    if (decode[enc] == -1) {
      decode[enc] = i;
    }
  }
}
inline int getnTime(int t) { return decode[encode(t)]; }
const int N = 1e6;
int t[60][N];
void update(int pos, int s, int e, int l, int val) {
  if (l < s || l > e) return;
  if (s == e) {
    for (int i = 0; i < 60; ++i) {
      t[i][pos] = 1 + (i % val == 0);
    }
    return;
  }
  int r = pos * 2;
  update(r, s, (s + e) / 2, l, val);
  update(r | 1, (s + e) / 2 + 1, e, l, val);
  for (int i = 0; i < 60; ++i) {
    s = t[i][r];
    e = t[getnTime(i + s)][r | 1];
    t[i][pos] = s + e;
  }
}
int query(int pos, int s, int e, int l, int r, int st) {
  if (e < l || s > r) return 0;
  if (s >= l && e <= r) {
    int nst = getnTime(st);
    return t[nst][pos];
  }
  int res = query(pos * 2, s, (s + e) / 2, l, r, st);
  res += query(pos * 2 + 1, (s + e) / 2 + 1, e, l, r, res + st);
  return res;
}
int main() {
  gen();
  int n, a, q, x;
  char c;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a);
    update(1, 1, n, i, a);
  }
  scanf("%d", &q);
  while (q--) {
    scanf(" %c %d %d", &c, &x, &a);
    if (c == 'C') {
      update(1, 1, n, x, a);
    } else {
      printf("%d\n", query(1, 1, n, x, a - 1, 0));
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int encode(int t) { return (t % 4) * 30 + (t % 5) * 6 + (t % 6); }
int decode[120];
void gen() {
  memset(decode, -1, sizeof decode);
  for (int i = 0; i < 120; ++i) {
    int enc = encode(i);
    if (decode[enc] == -1) {
      decode[enc] = i;
    }
  }
}
inline int getnTime(int t) { return decode[encode(t)]; }
const int N = 1e6;
int t[60][N];
void update(int pos, int s, int e, int l, int val) {
  if (l < s || l > e) return;
  if (s == e) {
    for (int i = 0; i < 60; ++i) {
      t[i][pos] = 1 + (i % val == 0);
    }
    return;
  }
  int r = pos * 2;
  update(r, s, (s + e) / 2, l, val);
  update(r | 1, (s + e) / 2 + 1, e, l, val);
  for (int i = 0; i < 60; ++i) {
    s = t[i][r];
    e = t[getnTime(i + s)][r | 1];
    t[i][pos] = s + e;
  }
}
int query(int pos, int s, int e, int l, int r, int st) {
  if (e < l || s > r) return 0;
  if (s >= l && e <= r) {
    int nst = getnTime(st);
    return t[nst][pos];
  }
  int res = query(pos * 2, s, (s + e) / 2, l, r, st);
  res += query(pos * 2 + 1, (s + e) / 2 + 1, e, l, r, res + st);
  return res;
}
int main() {
  gen();
  int n, a, q, x;
  char c;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a);
    update(1, 1, n, i, a);
  }
  scanf("%d", &q);
  while (q--) {
    scanf(" %c %d %d", &c, &x, &a);
    if (c == 'C') {
      update(1, 1, n, x, a);
    } else {
      printf("%d\n", query(1, 1, n, x, a - 1, 0));
    }
  }
  return 0;
}

```