#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3fffffff;
const long long mod = 1000000007;
const int maxn = 1e5 + 10;
int a[maxn];
int seg[60][maxn * 4];
void pushup(int x) {
  for (int i = 0; i < 60; i++) {
    seg[i][x] = seg[i][x * 2] + seg[(i + seg[i][x * 2]) % 60][x * 2 + 1];
  }
}
void build(int i, int l, int r) {
  if (l == r) {
    for (int o = 0; o < 60; o++) seg[o][i] = 1 + (o % a[l] == 0);
    return;
  }
  int m = (l + r) / 2;
  build(i * 2, l, m);
  build(i * 2 + 1, m + 1, r);
  pushup(i);
}
void update(int i, int l, int r, int pos, int v) {
  if (l == r) {
    for (int o = 0; o < 60; o++) seg[o][i] = 1 + (o % a[pos] == 0);
    return;
  }
  int m = (l + r) / 2;
  if (pos <= m)
    update(i * 2, l, m, pos, v);
  else
    update(i * 2 + 1, m + 1, r, pos, v);
  pushup(i);
}
int query(int o, int i, int L, int R, int l, int r) {
  if (L == l && R == r) {
    return seg[o][i];
  }
  int m = (L + R) / 2;
  if (r <= m)
    return query(o, i * 2, L, m, l, r);
  else if (l > m)
    return query(o, i * 2 + 1, m + 1, R, l, r);
  else {
    int ans = query(o, i * 2, L, m, l, m);
    return ans + query((o + ans) % 60, i * 2 + 1, m + 1, R, m + 1, r);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) scanf("%d", &a[i]);
  build(1, 1, n);
  int m;
  scanf("%d", &m);
  while (m--) {
    char s[2];
    scanf("%s", s);
    int x, y;
    scanf("%d%d", &x, &y);
    if (s[0] == 'C') {
      a[x] = y;
      update(1, 1, n, x, y);
    } else if (s[0] == 'A') {
      printf("%d\n", query(0, 1, 1, n, x, y - 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 int inf = 0x3fffffff;
const long long mod = 1000000007;
const int maxn = 1e5 + 10;
int a[maxn];
int seg[60][maxn * 4];
void pushup(int x) {
  for (int i = 0; i < 60; i++) {
    seg[i][x] = seg[i][x * 2] + seg[(i + seg[i][x * 2]) % 60][x * 2 + 1];
  }
}
void build(int i, int l, int r) {
  if (l == r) {
    for (int o = 0; o < 60; o++) seg[o][i] = 1 + (o % a[l] == 0);
    return;
  }
  int m = (l + r) / 2;
  build(i * 2, l, m);
  build(i * 2 + 1, m + 1, r);
  pushup(i);
}
void update(int i, int l, int r, int pos, int v) {
  if (l == r) {
    for (int o = 0; o < 60; o++) seg[o][i] = 1 + (o % a[pos] == 0);
    return;
  }
  int m = (l + r) / 2;
  if (pos <= m)
    update(i * 2, l, m, pos, v);
  else
    update(i * 2 + 1, m + 1, r, pos, v);
  pushup(i);
}
int query(int o, int i, int L, int R, int l, int r) {
  if (L == l && R == r) {
    return seg[o][i];
  }
  int m = (L + R) / 2;
  if (r <= m)
    return query(o, i * 2, L, m, l, r);
  else if (l > m)
    return query(o, i * 2 + 1, m + 1, R, l, r);
  else {
    int ans = query(o, i * 2, L, m, l, m);
    return ans + query((o + ans) % 60, i * 2 + 1, m + 1, R, m + 1, r);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) scanf("%d", &a[i]);
  build(1, 1, n);
  int m;
  scanf("%d", &m);
  while (m--) {
    char s[2];
    scanf("%s", s);
    int x, y;
    scanf("%d%d", &x, &y);
    if (s[0] == 'C') {
      a[x] = y;
      update(1, 1, n, x, y);
    } else if (s[0] == 'A') {
      printf("%d\n", query(0, 1, 1, n, x, y - 1));
    }
  }
  return 0;
}

```