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 <typename T>
void read(T &x);
template <typename T>
void write(T x);
template <typename T>
void writesp(T x);
template <typename T>
void writeln(T x);
const int N = 2e5 + 5;
int a[N];
namespace seg {
struct node {
  int l, r, val, t[61];
} tree[N << 2];
inline void pushup(int p) {
  for (register int i = 0; i < 60; i++) {
    int temp = tree[p << 1].t[i];
    int nxt = (i + temp) % 60;
    tree[p].t[i] = temp + tree[p << 1 | 1].t[nxt];
  }
}
inline void build(int p, int l, int r) {
  tree[p].l = l;
  tree[p].r = r;
  if (l == r) {
    for (register int i = 0; i < 60; i++) {
      if (i % a[l] == 0)
        tree[p].t[i] = 2;
      else
        tree[p].t[i] = 1;
    }
    return;
  }
  int mid = l + r >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  pushup(p);
}
inline void update(int p, int l, int r, int k, int val) {
  if (l == r) {
    a[l] = val;
    for (register int i = 0; i < 60; i++) {
      if (i % a[l] == 0)
        tree[p].t[i] = 2;
      else
        tree[p].t[i] = 1;
    }
    return;
  }
  int mid = l + r >> 1;
  if (k <= mid)
    update(p << 1, l, mid, k, val);
  else
    update(p << 1 | 1, mid + 1, r, k, val);
  pushup(p);
}
inline int query(int p, int u, int v, int t) {
  if (u <= tree[p].l && tree[p].r <= v) {
    return t + tree[p].t[t % 60];
  }
  int mid = tree[p].l + tree[p].r >> 1;
  if (u <= mid) t = query(p << 1, u, v, t);
  if (v > mid) t = query(p << 1 | 1, u, v, t);
  return t;
}
}  // namespace seg
int n, m;
int main() {
  read(n);
  for (register int i = 1; i <= n; i++) read(a[i]);
  seg::build(1, 1, n);
  read(m);
  char wn;
  int u, v;
  while (m--) {
    wn = getchar();
    while (wn != 'C' && wn != 'A') wn = getchar();
    read(u);
    read(v);
    if (wn == 'C') {
      seg::update(1, 1, n, u, v);
    } else
      writeln(seg::query(1, u, v - 1, 0));
  }
}
template <typename T>
void read(T &x) {
  char wn = getchar();
  int t = 1;
  x = 0;
  while (wn < '0' || wn > '9') {
    if (wn == '-') t = -1;
    wn = getchar();
  }
  while (wn >= '0' && wn <= '9') {
    x = x * 10 + wn - '0';
    wn = getchar();
  }
  x *= t;
}
template <typename T>
void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x <= 9) {
    putchar(x + '0');
    return;
  }
  write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  putchar('\n');
}
template <typename T>
void writesp(T x) {
  write(x);
  putchar(' ');
}
