#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long inf = 0xc0c0c0c0c0c0c0c0;
struct node {
  int time[60];
  node() { memset(time, 0, sizeof(time)); }
};
int N;
const int MAX = 1e5 + 7;
node tree[MAX << 2];
void get(int pos) {
  int pl = pos << 1;
  int pr = pos << 1 | 1;
  for (int i = 0; i < 60; i++) {
    int delaytime = tree[pl].time[i];
    delaytime += tree[pr].time[(delaytime + i) % 60];
    tree[pos].time[i] = delaytime;
  }
}
void query(int &TIME, int from, int to, int l, int r, int pos) {
  if (l >= from && r <= to) {
    TIME += tree[pos].time[TIME % 60];
    return;
  }
  int mid = (l + r) >> 1;
  if (mid >= from) query(TIME, from, to, l, mid, pos << 1);
  if (mid < to) query(TIME, from, to, mid + 1, r, pos << 1 | 1);
}
void update(int x, int val, int l, int r, int pos) {
  if (x == l && x == r) {
    for (int i = 0; i < 60; i++) tree[pos].time[i] = (i % val) ? 1 : 2;
    return;
  }
  int mid = (l + r) >> 1;
  if (mid >= x)
    update(x, val, l, mid, pos << 1);
  else
    update(x, val, mid + 1, r, pos << 1 | 1);
  get(pos);
}
void init(int l, int r, int pos) {
  if (l == r) {
    int x;
    scanf("%d", &x);
    for (int i = 0; i < 60; i++) tree[pos].time[i] = (i % x) ? 1 : 2;
    return;
  }
  int mid = (l + r) >> 1;
  init(l, mid, pos << 1);
  init(mid + 1, r, pos << 1 | 1);
  get(pos);
}
int main() {
  scanf("%d", &N);
  init(1, N, 1);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    char act[3];
    int l, r;
    scanf("%s%d%d", act, &l, &r);
    if (act[0] == 'A') {
      int ans = 0;
      query(ans, l, r - 1, 1, N, 1);
      printf("%d\n", ans);
    } else {
      update(l, r, 1, N, 1);
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 double pi = acos(-1);
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long inf = 0xc0c0c0c0c0c0c0c0;
struct node {
  int time[60];
  node() { memset(time, 0, sizeof(time)); }
};
int N;
const int MAX = 1e5 + 7;
node tree[MAX << 2];
void get(int pos) {
  int pl = pos << 1;
  int pr = pos << 1 | 1;
  for (int i = 0; i < 60; i++) {
    int delaytime = tree[pl].time[i];
    delaytime += tree[pr].time[(delaytime + i) % 60];
    tree[pos].time[i] = delaytime;
  }
}
void query(int &TIME, int from, int to, int l, int r, int pos) {
  if (l >= from && r <= to) {
    TIME += tree[pos].time[TIME % 60];
    return;
  }
  int mid = (l + r) >> 1;
  if (mid >= from) query(TIME, from, to, l, mid, pos << 1);
  if (mid < to) query(TIME, from, to, mid + 1, r, pos << 1 | 1);
}
void update(int x, int val, int l, int r, int pos) {
  if (x == l && x == r) {
    for (int i = 0; i < 60; i++) tree[pos].time[i] = (i % val) ? 1 : 2;
    return;
  }
  int mid = (l + r) >> 1;
  if (mid >= x)
    update(x, val, l, mid, pos << 1);
  else
    update(x, val, mid + 1, r, pos << 1 | 1);
  get(pos);
}
void init(int l, int r, int pos) {
  if (l == r) {
    int x;
    scanf("%d", &x);
    for (int i = 0; i < 60; i++) tree[pos].time[i] = (i % x) ? 1 : 2;
    return;
  }
  int mid = (l + r) >> 1;
  init(l, mid, pos << 1);
  init(mid + 1, r, pos << 1 | 1);
  get(pos);
}
int main() {
  scanf("%d", &N);
  init(1, N, 1);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    char act[3];
    int l, r;
    scanf("%s%d%d", act, &l, &r);
    if (act[0] == 'A') {
      int ans = 0;
      query(ans, l, r - 1, 1, N, 1);
      printf("%d\n", ans);
    } else {
      update(l, r, 1, N, 1);
    }
  }
  return 0;
}

```