#include <bits/stdc++.h>
using namespace std;
inline bool isvowel(char c) {
  c = tolower(c);
  if (c == 'a' || c == 'e' || c == 'i' || c == 'y' || c == 'o' || c == 'u')
    return 1;
  return 0;
}
const double eps = 0.000001;
const long double pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const long long MOD = 1e18 + 9;
const long long INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 1e5 + 123;
const int M = 22;
const int pri = 997;
const int Magic = 2101;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
int n, m, k;
int c[N];
struct node {
  int a[60];
  node() {
    for (int i = 0; i < 60; i++) a[i] = 0;
  }
} t[N * 4];
node merge(node a, node b) {
  node res;
  for (int i = 0; i < 60; i++) {
    res.a[i] = a.a[i] + b.a[(i + a.a[i]) % 60];
  }
  return res;
}
void build(int v = 1, int tl = 1, int tr = n) {
  if (tl == tr) {
    for (int i = 0; i < 60; i++) {
      t[v].a[i] = 1;
      if (i % c[tl] == 0) {
        t[v].a[i]++;
      }
    }
    return;
  }
  int tm = (tl + tr) >> 1;
  build((v << 1), tl, tm);
  build((v << 1) | 1, tm + 1, tr);
  t[v] = merge(t[(v << 1)], t[(v << 1) | 1]);
}
int get(int l, int r, int start = 0, int v = 1, int tl = 1, int tr = n) {
  if (tl > r || tr < l) return 0;
  if (tl >= l && tr <= r) {
    return t[v].a[start];
  }
  int tm = (tl + tr) >> 1;
  int x = get(l, r, start, (v << 1), tl, tm);
  int y = get(l, r, (start + x) % 60, (v << 1) | 1, tm + 1, tr);
  return x + y;
}
void upd(int pos, int x, int v = 1, int tl = 1, int tr = n) {
  if (tl == tr) {
    for (int i = 0; i < 60; i++) {
      t[v].a[i] = 1;
      if (i % x == 0) {
        t[v].a[i]++;
      }
    }
    return;
  }
  int tm = (tl + tr) >> 1;
  if (pos <= tm)
    upd(pos, x, (v << 1), tl, tm);
  else
    upd(pos, x, (v << 1) | 1, tm + 1, tr);
  t[v] = merge(t[(v << 1)], t[(v << 1) | 1]);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
  }
  build();
  int q;
  cin >> q;
  while (q--) {
    char tp;
    int x, y;
    cin >> tp >> x >> y;
    if (tp == 'A') {
      y--;
      cout << get(x, y) << '\n';
      continue;
    }
    upd(x, y);
  }
  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;
inline bool isvowel(char c) {
  c = tolower(c);
  if (c == 'a' || c == 'e' || c == 'i' || c == 'y' || c == 'o' || c == 'u')
    return 1;
  return 0;
}
const double eps = 0.000001;
const long double pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const long long MOD = 1e18 + 9;
const long long INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 1e5 + 123;
const int M = 22;
const int pri = 997;
const int Magic = 2101;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
int n, m, k;
int c[N];
struct node {
  int a[60];
  node() {
    for (int i = 0; i < 60; i++) a[i] = 0;
  }
} t[N * 4];
node merge(node a, node b) {
  node res;
  for (int i = 0; i < 60; i++) {
    res.a[i] = a.a[i] + b.a[(i + a.a[i]) % 60];
  }
  return res;
}
void build(int v = 1, int tl = 1, int tr = n) {
  if (tl == tr) {
    for (int i = 0; i < 60; i++) {
      t[v].a[i] = 1;
      if (i % c[tl] == 0) {
        t[v].a[i]++;
      }
    }
    return;
  }
  int tm = (tl + tr) >> 1;
  build((v << 1), tl, tm);
  build((v << 1) | 1, tm + 1, tr);
  t[v] = merge(t[(v << 1)], t[(v << 1) | 1]);
}
int get(int l, int r, int start = 0, int v = 1, int tl = 1, int tr = n) {
  if (tl > r || tr < l) return 0;
  if (tl >= l && tr <= r) {
    return t[v].a[start];
  }
  int tm = (tl + tr) >> 1;
  int x = get(l, r, start, (v << 1), tl, tm);
  int y = get(l, r, (start + x) % 60, (v << 1) | 1, tm + 1, tr);
  return x + y;
}
void upd(int pos, int x, int v = 1, int tl = 1, int tr = n) {
  if (tl == tr) {
    for (int i = 0; i < 60; i++) {
      t[v].a[i] = 1;
      if (i % x == 0) {
        t[v].a[i]++;
      }
    }
    return;
  }
  int tm = (tl + tr) >> 1;
  if (pos <= tm)
    upd(pos, x, (v << 1), tl, tm);
  else
    upd(pos, x, (v << 1) | 1, tm + 1, tr);
  t[v] = merge(t[(v << 1)], t[(v << 1) | 1]);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
  }
  build();
  int q;
  cin >> q;
  while (q--) {
    char tp;
    int x, y;
    cin >> tp >> x >> y;
    if (tp == 'A') {
      y--;
      cout << get(x, y) << '\n';
      continue;
    }
    upd(x, y);
  }
  return 0;
}

```