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 = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e5 + 5;
inline long long read() {
  long long ans = 0;
  char ch = getchar(), last = ' ';
  while (!isdigit(ch)) {
    last = ch;
    ch = getchar();
  }
  while (isdigit(ch)) {
    ans = ans * 10 + ch - '0';
    ch = getchar();
  }
  if (last == '-') ans = -ans;
  return ans;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
int n, m;
char s[2];
struct Tree {
  int l, r;
  int tim[65];
} t[maxn << 2];
void pushup(int now) {
  for (int i = 0; i < 60; ++i)
    t[now].tim[i] =
        t[now << 1].tim[i] + t[now << 1 | 1].tim[(i + t[now << 1].tim[i]) % 60];
}
void build(int L, int R, int now) {
  t[now].l = L;
  t[now].r = R;
  if (L == R) {
    int x = read();
    for (int i = 0; i < 60; ++i)
      if (!(i % x))
        t[now].tim[i] = 2;
      else
        t[now].tim[i] = 1;
    return;
  }
  int mid = (L + R) >> 1;
  build(L, mid, now << 1);
  build(mid + 1, R, now << 1 | 1);
  pushup(now);
}
void update(int idx, int d, int now) {
  if (t[now].l == t[now].r) {
    for (int i = 0; i < 60; ++i)
      if (!(i % d))
        t[now].tim[i] = 2;
      else
        t[now].tim[i] = 1;
    return;
  }
  int mid = (t[now].l + t[now].r) >> 1;
  if (idx <= mid)
    update(idx, d, now << 1);
  else
    update(idx, d, now << 1 | 1);
  pushup(now);
}
int query(int L, int R, int now, int Tim) {
  if (t[now].l == L && t[now].r == R) return t[now].tim[Tim % 60];
  int mid = (t[now].l + t[now].r) >> 1;
  if (R <= mid)
    return query(L, R, now << 1, Tim);
  else if (L > mid)
    return query(L, R, now << 1 | 1, Tim);
  else {
    int ret = query(L, mid, now << 1, Tim);
    ret += query(mid + 1, R, now << 1 | 1, (Tim + ret) % 60);
    return ret;
  }
}
int main() {
  n = read();
  build(1, n, 1);
  m = read();
  for (int i = 1; i <= m; ++i) {
    scanf("%s", s);
    if (s[0] == 'C') {
      int x = read(), d = read();
      update(x, d, 1);
    } else {
      int L = read(), R = read();
      write(query(L, R - 1, 1, 0));
      puts("");
    }
  }
  return 0;
}
