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;
template <class T>
inline T min(T &a, T &b) {
  return a < b ? a : b;
}
template <class T>
inline T max(T &a, T &b) {
  return a > b ? a : b;
}
template <class T>
void read(T &x) {
  char ch;
  while ((ch = getchar()) && !(ch >= '0' && ch <= '9'))
    ;
  x = ch - '0';
  while ((ch = getchar()) && (ch >= '0' && ch <= '9')) x = x * 10 + ch - '0';
}
struct point {
  int x, y;
  point() {}
  point(int _x, int _y) : x(_x), y(_y) {}
};
long long Pow(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long Pow(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    a = a * a;
  }
  return res;
}
const int N = 120000;
int n, L[N], R[N], a[N], pos[N], Len, flag[N];
struct lj {
  int l, r, val[60];
} t[N << 2];
void update(int w) {
  if (t[w].l == t[w].r) return;
  for (int i = 0; i <= 59; i++)
    t[w].val[i] =
        t[w << 1].val[i] + t[w << 1 | 1].val[(i + t[w << 1].val[i]) % 60];
}
void build(int w, int l, int r) {
  t[w].l = l;
  t[w].r = r;
  if (l == r) {
    for (int i = 0; i <= 59; i++)
      if (i % a[l] == 0)
        t[w].val[i] = 2;
      else
        t[w].val[i] = 1;
    return;
  }
  int mid = l + r >> 1;
  build(w << 1, l, mid);
  build(w << 1 | 1, mid + 1, r);
  update(w);
}
void change(int w, int l, int c) {
  if (t[w].l == t[w].r) {
    for (int i = 0; i <= 59; i++)
      if (i % a[l] == 0)
        t[w].val[i] = 2;
      else
        t[w].val[i] = 1;
    return;
  }
  int mid = t[w].l + t[w].r >> 1;
  if (l <= mid)
    change(w << 1, l, c);
  else
    change(w << 1 | 1, l, c);
  update(w);
}
int Go(int w, int l, int r, int c) {
  if (t[w].l == l && t[w].r == r) return t[w].val[c];
  int mid = t[w].l + t[w].r >> 1;
  if (r <= mid)
    return Go(w << 1, l, r, c);
  else if (l > mid)
    return Go(w << 1 | 1, l, r, c);
  else {
    int d = Go(w << 1, l, mid, c);
    return d + Go(w << 1 | 1, mid + 1, r, (c + d) % 60);
  }
}
char s[3];
int main() {
  int q, x, y;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  n++;
  a[n] = 123;
  scanf("%d", &q);
  build(1, 1, n);
  while (q--) {
    scanf("%s%d%d", s, &x, &y);
    if (s[0] == 'C')
      a[x] = y, change(1, x, y);
    else if (s[0] == 'A')
      printf("%d\n", Go(1, x, y - 1, 0));
  }
}
