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;
inline char gc() {
  static char now[1 << 16], *S, *T;
  if (T == S) {
    T = (S = now) + fread(now, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline long long read() {
  register long long x = 0, f = 1;
  char ch = gc();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = gc();
  }
  return (f == 1) ? x : -x;
}
struct node {
  int mx[60];
} seg[400011];
inline node operator+(const node &a, const node &b) {
  node c;
  for (int i = 0; i < 60; i++) {
    c.mx[i] = a.mx[i] + b.mx[(i + a.mx[i]) % 60];
  }
  return c;
}
inline void pushup(int rt) { seg[rt] = seg[rt << 1] + seg[rt << 1 | 1]; }
int n, a[200001];
void build(int rt, int lb, int rb) {
  if (lb == rb) {
    for (int i = 0; i < 60; i++) seg[rt].mx[i] = ((i % a[lb]) ? 1 : 2);
    return;
  }
  build(rt << 1, lb, ((lb + rb) >> 1));
  build(rt << 1 | 1, ((lb + rb) >> 1) + 1, rb);
  pushup(rt);
}
void update(int rt, int lb, int rb, int pos) {
  if (lb == rb) {
    for (int i = 0; i < 60; i++) seg[rt].mx[i] = ((i % a[lb]) ? 1 : 2);
    return;
  }
  (((lb + rb) >> 1) >= pos
       ? update(rt << 1, lb, ((lb + rb) >> 1), pos)
       : update(rt << 1 | 1, ((lb + rb) >> 1) + 1, rb, pos));
  pushup(rt);
}
node ANS;
void query(int rt, int lb, int rb, int l, int r) {
  if (lb > r || rb < l) return;
  if (lb >= l && rb <= r) return (void)(ANS = ANS + seg[rt]);
  query(rt << 1, lb, ((lb + rb) >> 1), l, r);
  query(rt << 1 | 1, ((lb + rb) >> 1) + 1, rb, l, r);
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
  }
  build(1, 1, n);
  int Q = read();
  while (Q--) {
    char s = gc();
    while (!isalpha(s)) s = gc();
    if (s == 'C') {
      int pos = read(), x = read();
      a[pos] = x;
      update(1, 1, n, pos);
    } else {
      int a = read(), b = read();
      if (a == b) {
        printf("0\n");
        continue;
      } else {
        b--;
        for (int i = 0; i < 60; i++) ANS.mx[i] = 0;
        query(1, 1, n, a, b);
        printf("%d\n", ANS.mx[0]);
      }
    }
  }
  return 0;
}
