#include <bits/stdc++.h>
using namespace std;
inline void read(int &x) {
  char ch;
  int t = 0;
  x = 0;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') t = 1;
  for (; ch >= '0' && ch <= '9'; x = x * 10 + ch - 48, ch = getchar())
    ;
  if (t) x = -x;
}
const int N = 100005;
int sum[N << 2][61], a[N], b[60][6];
void pushup(int rt) {
  for (int i = 0; i < 60; i++)
    sum[rt][i] = sum[rt << 1][i] + sum[rt << 1 | 1][(i + sum[rt << 1][i]) % 60];
}
void build(int l, int r, int rt) {
  if (l == r) {
    for (int i = 0; i < 60; i++) sum[rt][i] = (i % a[l] == 0) ? 2 : 1;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  pushup(rt);
}
void update(int p, int c, int l, int r, int rt) {
  if (l == r) {
    for (int i = 0; i < 60; i++) sum[rt][i] = (i % c == 0) ? 2 : 1;
    return;
  }
  int m = (l + r) >> 1;
  if (p <= m)
    update(p, c, l, m, rt << 1);
  else
    update(p, c, m + 1, r, rt << 1 | 1);
  pushup(rt);
}
int query(int L, int R, int p, int l, int r, int rt) {
  if (L <= l && r <= R) return sum[rt][p];
  int m = (l + r) >> 1;
  if (L <= m && m < R) {
    int t = query(L, R, p, l, m, rt << 1);
    return t + query(L, R, (p + t) % 60, m + 1, r, rt << 1 | 1);
  }
  if (L <= m) return query(L, R, p, l, m, rt << 1);
  if (m < R) return query(L, R, p, m + 1, r, rt << 1 | 1);
}
char op[3];
int main() {
  int n, i, j, k, l, r, p, m, T;
  int t, x, y, ans;
  for (i = 0; i < 60; i++)
    for (b[i][0] = 2, j = 1; j < 6; j++) b[i][j] = (i % j == 0) ? 2 : 1;
  while (~scanf("%d", &n)) {
    for (i = 1; i <= n; i++) read(a[i]);
    build(1, n, 1);
    scanf("%d", &m);
    while (m--) {
      scanf("%s", op);
      read(x), read(y);
      if (op[0] == 'C')
        update(x, y, 1, n, 1);
      else
        printf("%d\n", query(x, y - 1, 0, 1, n, 1));
    }
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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>
using namespace std;
inline void read(int &x) {
  char ch;
  int t = 0;
  x = 0;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') t = 1;
  for (; ch >= '0' && ch <= '9'; x = x * 10 + ch - 48, ch = getchar())
    ;
  if (t) x = -x;
}
const int N = 100005;
int sum[N << 2][61], a[N], b[60][6];
void pushup(int rt) {
  for (int i = 0; i < 60; i++)
    sum[rt][i] = sum[rt << 1][i] + sum[rt << 1 | 1][(i + sum[rt << 1][i]) % 60];
}
void build(int l, int r, int rt) {
  if (l == r) {
    for (int i = 0; i < 60; i++) sum[rt][i] = (i % a[l] == 0) ? 2 : 1;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  pushup(rt);
}
void update(int p, int c, int l, int r, int rt) {
  if (l == r) {
    for (int i = 0; i < 60; i++) sum[rt][i] = (i % c == 0) ? 2 : 1;
    return;
  }
  int m = (l + r) >> 1;
  if (p <= m)
    update(p, c, l, m, rt << 1);
  else
    update(p, c, m + 1, r, rt << 1 | 1);
  pushup(rt);
}
int query(int L, int R, int p, int l, int r, int rt) {
  if (L <= l && r <= R) return sum[rt][p];
  int m = (l + r) >> 1;
  if (L <= m && m < R) {
    int t = query(L, R, p, l, m, rt << 1);
    return t + query(L, R, (p + t) % 60, m + 1, r, rt << 1 | 1);
  }
  if (L <= m) return query(L, R, p, l, m, rt << 1);
  if (m < R) return query(L, R, p, m + 1, r, rt << 1 | 1);
}
char op[3];
int main() {
  int n, i, j, k, l, r, p, m, T;
  int t, x, y, ans;
  for (i = 0; i < 60; i++)
    for (b[i][0] = 2, j = 1; j < 6; j++) b[i][j] = (i % j == 0) ? 2 : 1;
  while (~scanf("%d", &n)) {
    for (i = 1; i <= n; i++) read(a[i]);
    build(1, n, 1);
    scanf("%d", &m);
    while (m--) {
      scanf("%s", op);
      read(x), read(y);
      if (op[0] == 'C')
        update(x, y, 1, n, 1);
      else
        printf("%d\n", query(x, y - 1, 0, 1, n, 1));
    }
  }
  return 0;
}

```