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;
struct node {
  int len[60];
  int to[60];
  node() {
    for (int i = 0; i < (60); i++) {
      len[i] = 0;
      to[i] = i;
    }
  }
  node merge(const node &b) {
    node res;
    for (int i = 0; i < (60); i++) {
      res.to[i] = b.to[to[i]];
      res.len[i] = b.len[to[i]] + len[i];
    }
    return res;
  }
  void makeNode(int a) {
    for (int i = 0; i < 60; i++) {
      if (i % a == 0) {
        to[i] = (i + 2) % 60;
        len[i] = 2;
      } else {
        to[i] = (i + 1) % 60;
        len[i] = 1;
      }
    }
  }
};
const int MAXN = 131072 * 2;
struct SegmentTree {
  node nd[MAXN];
  int ds;
  static int getds(int n) {
    n--;
    n |= n >> 1;
    n |= n >> 2;
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
    return n + 1;
  }
  SegmentTree(int n, int *a) {
    ds = getds(n);
    for (int i = 0; i < (n); i++) nd[i + ds].makeNode(a[i]);
    for (int i = ds - 1; i > 0; i--) nd[i] = nd[2 * i].merge(nd[2 * i + 1]);
  }
  node getRes(int l, int r) { return getRes(1, 0, ds - 1, l, r); }
  node getRes(int cur, int cl, int cr, int l, int r) {
    if (l <= cl && cr <= r)
      return nd[cur];
    else if (r < cl || cr < l)
      return node();
    else {
      int mid = (cl + cr) / 2;
      node lRes = getRes(2 * cur, cl, mid, l, r),
           rRes = getRes(2 * cur + 1, mid + 1, cr, l, r);
      return lRes.merge(rRes);
    }
  }
  void update(int i, int a) {
    i += ds;
    nd[i].makeNode(a);
    i >>= 1;
    while (i) {
      nd[i] = nd[2 * i].merge(nd[2 * i + 1]);
      i >>= 1;
    }
  }
};
int a[MAXN];
SegmentTree *s;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) scanf("%d", &a[i]);
  s = new SegmentTree(n, a);
  int q;
  scanf("%d", &q);
  while (q--) {
    char buf[10];
    scanf("%s", buf);
    if (buf[0] == 'A') {
      int l, r;
      scanf("%d%d", &l, &r);
      l--;
      r -= 2;
      node res = s->getRes(l, r);
      printf("%d\n", res.len[0]);
    } else {
      int x, a;
      scanf("%d%d", &x, &a);
      x--;
      s->update(x, a);
    }
  }
  return 0;
}
