#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 3;
int n, q;
int a[MAXN];
void read() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  cin >> q;
}
int tr[MAXN * 4][60];
void merge(int v) {
  for (int i = 0; i <= 59; i++)
    tr[v][i] = tr[v * 2][i] + tr[v * 2 + 1][(i + tr[v * 2][i]) % 60];
}
void build(int v, int l, int r) {
  if (l == r) {
    for (int i = 0; i <= 59; i++) {
      if (i % a[l])
        tr[v][i] = 1;
      else
        tr[v][i] = 2;
    }
    return;
  }
  int mid = (l + r) / 2;
  build(v * 2, l, mid);
  build(v * 2 + 1, mid + 1, r);
  merge(v);
}
void update(int v, int l, int r, int pos, int val) {
  if (l == r) {
    for (int i = 0; i <= 59; i++) {
      if (i % val)
        tr[v][i] = 1;
      else
        tr[v][i] = 2;
    }
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(v * 2, l, mid, pos, val);
  else
    update(v * 2 + 1, mid + 1, r, pos, val);
  merge(v);
}
int ans_time;
void get(int v, int l, int r, int ll, int rr) {
  if (r < ll || l > rr) return;
  if (l >= ll && r <= rr) {
    ans_time += tr[v][ans_time % 60];
    return;
  }
  int mid = (l + r) / 2;
  get(v * 2, l, mid, ll, rr);
  get(v * 2 + 1, mid + 1, r, ll, rr);
}
void solve() {
  build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    char type;
    int x, y;
    cin >> type >> x >> y;
    if (type == 'A') {
      ans_time = 0;
      get(1, 1, n, x, y - 1);
      cout << ans_time << endl;
    } else
      update(1, 1, n, x, y);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  read();
  solve();
}

### Prompt

In CPP, your task is to solve 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>
using namespace std;
const int MAXN = 1e5 + 3;
int n, q;
int a[MAXN];
void read() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  cin >> q;
}
int tr[MAXN * 4][60];
void merge(int v) {
  for (int i = 0; i <= 59; i++)
    tr[v][i] = tr[v * 2][i] + tr[v * 2 + 1][(i + tr[v * 2][i]) % 60];
}
void build(int v, int l, int r) {
  if (l == r) {
    for (int i = 0; i <= 59; i++) {
      if (i % a[l])
        tr[v][i] = 1;
      else
        tr[v][i] = 2;
    }
    return;
  }
  int mid = (l + r) / 2;
  build(v * 2, l, mid);
  build(v * 2 + 1, mid + 1, r);
  merge(v);
}
void update(int v, int l, int r, int pos, int val) {
  if (l == r) {
    for (int i = 0; i <= 59; i++) {
      if (i % val)
        tr[v][i] = 1;
      else
        tr[v][i] = 2;
    }
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(v * 2, l, mid, pos, val);
  else
    update(v * 2 + 1, mid + 1, r, pos, val);
  merge(v);
}
int ans_time;
void get(int v, int l, int r, int ll, int rr) {
  if (r < ll || l > rr) return;
  if (l >= ll && r <= rr) {
    ans_time += tr[v][ans_time % 60];
    return;
  }
  int mid = (l + r) / 2;
  get(v * 2, l, mid, ll, rr);
  get(v * 2 + 1, mid + 1, r, ll, rr);
}
void solve() {
  build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    char type;
    int x, y;
    cin >> type >> x >> y;
    if (type == 'A') {
      ans_time = 0;
      get(1, 1, n, x, y - 1);
      cout << ans_time << endl;
    } else
      update(1, 1, n, x, y);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  read();
  solve();
}

```