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
#include <bits/stdc++.h>
using namespace std;
const int inf = ~0U >> 1;
const long long INF = ~0ULL >> 1;
template <class T>
inline void read(T &n) {
  char c;
  for (c = getchar(); !(c >= '0' && c <= '9'); c = getchar())
    ;
  n = c - '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0';
}
int Pow(int base, int n, int mo) {
  if (n == 0) return 1;
  if (n == 1) return base % mo;
  int tmp = Pow(base, n >> 1, mo);
  tmp = (long long)tmp * tmp % mo;
  if (n & 1) tmp = (long long)tmp * base % mo;
  return tmp;
}
struct node {
  int l, r;
  int d[65];
} t[120000 * 5];
int a[120000];
int N, M;
void build(int p, int l, int r) {
  t[p].l = l;
  t[p].r = r;
  if (l != r) {
    build(p << 1, l, l + r >> 1);
    build(p << 1 | 1, (l + r >> 1) + 1, r);
    for (int k = (0); k <= (59); ++k) {
      t[p].d[k] = t[p << 1].d[k];
      t[p].d[k] += t[p << 1 | 1].d[(k + t[p].d[k]) % 60];
    }
  } else {
    for (int k = (0); k <= (59); ++k)
      if (k % a[l] == 0)
        t[p].d[k] = 2;
      else
        t[p].d[k] = 1;
  }
}
void modify(int p, int dest) {
  if (t[p].l == t[p].r) {
    for (int k = (0); k <= (59); ++k)
      if (k % a[dest] == 0)
        t[p].d[k] = 2;
      else
        t[p].d[k] = 1;
    return;
  }
  int m = t[p].l + t[p].r >> 1;
  if (dest <= m)
    modify(p << 1, dest);
  else
    modify(p << 1 | 1, dest);
  for (int k = (0); k <= (59); ++k) {
    t[p].d[k] = t[p << 1].d[k];
    t[p].d[k] += t[p << 1 | 1].d[(k + t[p].d[k]) % 60];
  }
}
int query(int p, int l, int r, int cur) {
  if (t[p].l == l && t[p].r == r) return t[p].d[cur];
  int m = t[p].l + t[p].r >> 1;
  if (r <= m)
    return query(p << 1, l, r, cur);
  else if (l > m)
    return query(p << 1 | 1, l, r, cur);
  else {
    int ret = query(p << 1, l, m, cur);
    ret += query(p << 1 | 1, m + 1, r, (cur + ret) % 60);
    return ret;
  }
}
int main() {
  scanf("%d", &N);
  for (int i = (1); i <= (N); ++i) scanf("%d", &a[i]);
  build(1, 1, N);
  scanf("%d", &M);
  while (M--) {
    char ty[5];
    int x, y;
    scanf("%s", ty);
    if (ty[0] == 'C') {
      scanf("%d%d", &x, &y);
      a[x] = y;
      modify(1, x);
    } else {
      scanf("%d%d", &x, &y);
      printf("%d\n", query(1, x, y - 1, 0));
    }
  }
  return 0;
}
