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 re() {
  T N = 0;
  char c = getchar();
  bool neg = 0;
  for (; c < '0' || c > '9'; c = getchar()) neg |= c == '-';
  for (; c >= '0' && c <= '9'; c = getchar()) N = (N << 3) + (N << 1) + c - '0';
  return neg ? -N : N;
}
const int MX = 1e5;
int n, q;
int a[MX + 5];
struct Node {
  int res[60], end;
  Node(int e = 0x3f3f3f3f) : end(e) {
    for (int i = 0; i < 60; i++) res[i] = (i % end ? 1 : 2);
  }
  Node operator+(const Node &rhs) const {
    Node ret;
    for (int i = 0; i < 60; i++) {
      int newtime = res[i];
      ret.res[i] = newtime + rhs.res[(i + newtime) % 60];
    }
    ret.end = rhs.end;
    return ret;
  }
} st[MX * 4 + 10];
void build(int p = 1, int l = 1, int r = n) {
  if (l == r) return void(st[p] = Node(a[l]));
  int md = (l + r) >> 1;
  build(p << 1, l, md);
  build(p << 1 | 1, md + 1, r);
  st[p] = st[p << 1] + st[p << 1 | 1];
}
void upd(int i, int x, int p = 1, int l = 1, int r = n) {
  if (l == r) {
    a[l] = x;
    st[p] = Node(x);
    return;
  }
  int md = (l + r) >> 1;
  if (i <= md)
    upd(i, x, p << 1, l, md);
  else
    upd(i, x, p << 1 | 1, md + 1, r);
  st[p] = st[p << 1] + st[p << 1 | 1];
}
int query(int i, int j, int p = 1, int l = 1, int r = n, int cur = 0) {
  if (i > j || j < l || i > r) return 0;
  if (i <= l && j >= r) return st[p].res[cur];
  int md = (l + r) >> 1;
  int froml = query(i, j, p << 1, l, md, cur);
  int fromr = query(i, j, p << 1 | 1, md + 1, r, (cur + froml) % 60);
  return froml + fromr;
}
inline void init() { build(); }
inline int get(int l, int r) { return query(l, r - 1); }
void setval(int pos, int v) { upd(pos, v); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  init();
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    char buf[5];
    int x, y;
    scanf("%s %d %d", buf, &x, &y);
    if (buf[0] == 'C') {
      setval(x, y);
    } else {
      printf("%d\n", get(x, y));
    }
  }
  return 0;
}
