#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
const int MOD = 1000000007;
const int MAXN = 100010;
int a[MAXN], sum[60][MAXN * 4];
int n, q;
void pushup(int x) {
  for (int i = 0; i < 60; i++)
    sum[i][x] = sum[i][x + x] + sum[(i + sum[i][x + x]) % 60][x + x + 1];
}
void build(int x, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 60; i++) sum[i][x] = 1 + (i % a[l] == 0);
    return;
  }
  int y = (l + r) >> 1;
  build(x + x, l, y);
  build(x + x + 1, y + 1, r);
  pushup(x);
}
void updat(int x, int l, int r, int ll, int val) {
  if (ll < l || r < ll) return;
  if (l == r) {
    a[l] = val;
    for (int i = 0; i < 60; i++) sum[i][x] = 1 + (i % a[l] == 0);
    return;
  }
  int y = (l + r) >> 1;
  updat(x + x, l, y, ll, val);
  updat(x + x + 1, y + 1, r, ll, val);
  pushup(x);
}
int query(int x, int l, int r, int ll, int rr, int t) {
  if (rr < l || r < ll) return 0;
  if (ll <= l && r <= rr) return sum[t % 60][x];
  int y = (l + r) >> 1;
  int tmp = query(x + x, l, y, ll, rr, t);
  return tmp + query(x + x + 1, y + 1, r, ll, rr, tmp + t);
}
void solve() {
  char op[5];
  int ll, rr;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  build(1, 0, n - 1);
  scanf("%d", &q);
  while (q--) {
    scanf("%s%d%d", op, &ll, &rr);
    if (op[0] == 'C') {
      ll--;
      updat(1, 0, n - 1, ll, rr);
    } else {
      ll--;
      rr--;
      printf("%d\n", query(1, 0, n - 1, ll, rr - 1, 0));
    }
  }
}
int main() {
  solve();
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

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>
#pragma warning(disable : 4996)
using namespace std;
const int MOD = 1000000007;
const int MAXN = 100010;
int a[MAXN], sum[60][MAXN * 4];
int n, q;
void pushup(int x) {
  for (int i = 0; i < 60; i++)
    sum[i][x] = sum[i][x + x] + sum[(i + sum[i][x + x]) % 60][x + x + 1];
}
void build(int x, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 60; i++) sum[i][x] = 1 + (i % a[l] == 0);
    return;
  }
  int y = (l + r) >> 1;
  build(x + x, l, y);
  build(x + x + 1, y + 1, r);
  pushup(x);
}
void updat(int x, int l, int r, int ll, int val) {
  if (ll < l || r < ll) return;
  if (l == r) {
    a[l] = val;
    for (int i = 0; i < 60; i++) sum[i][x] = 1 + (i % a[l] == 0);
    return;
  }
  int y = (l + r) >> 1;
  updat(x + x, l, y, ll, val);
  updat(x + x + 1, y + 1, r, ll, val);
  pushup(x);
}
int query(int x, int l, int r, int ll, int rr, int t) {
  if (rr < l || r < ll) return 0;
  if (ll <= l && r <= rr) return sum[t % 60][x];
  int y = (l + r) >> 1;
  int tmp = query(x + x, l, y, ll, rr, t);
  return tmp + query(x + x + 1, y + 1, r, ll, rr, tmp + t);
}
void solve() {
  char op[5];
  int ll, rr;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  build(1, 0, n - 1);
  scanf("%d", &q);
  while (q--) {
    scanf("%s%d%d", op, &ll, &rr);
    if (op[0] == 'C') {
      ll--;
      updat(1, 0, n - 1, ll, rr);
    } else {
      ll--;
      rr--;
      printf("%d\n", query(1, 0, n - 1, ll, rr - 1, 0));
    }
  }
}
int main() {
  solve();
  return 0;
}

```