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>
const int vmax = 500000;
const int omax = 60;
using namespace std;
struct node {
  int ost[omax];
};
node v[vmax] = {0};
int n, m, x, y, p;
char c;
int build_tree(int n) {
  int j = 1;
  while (j < n) j *= 2;
  return j;
}
void makebase(int vn, int period) {
  int cur = 1;
  v[vn].ost[0] = 2;
  for (int i = 1; i < omax; ++i) {
    v[vn].ost[i] = (cur < period) ? 1 : 2;
    if (cur == period) cur = 0;
    cur++;
  }
}
void recount(int vn) {
  int phase;
  for (int i = 0; i < omax; ++i) {
    v[vn].ost[i] = v[2 * vn].ost[i];
    phase = (i + v[vn].ost[i]) % omax;
    v[vn].ost[i] += v[2 * vn + 1].ost[phase];
  }
}
void renew(int vn, int period) {
  makebase(vn, period);
  while ((vn >>= 1) > 0) recount(vn);
}
int find(int l, int r, int curt) {
  if (l > r) return 0;
  if (l == r) return v[l].ost[curt];
  int ans = 0;
  if ((l & 1) == 1) {
    ans += v[l].ost[curt];
    curt = (curt + v[l].ost[curt]) % omax;
    l++;
  }
  if ((r & 1) == 0) {
    int pres = find(l >> 1, (r - 1) >> 1, curt);
    ans += pres;
    curt = (curt + pres) % omax;
    ans += v[r].ost[curt];
  } else
    ans += find(l >> 1, r >> 1, curt);
  return ans;
}
int main(int argc, const char* argv[]) {
  cin >> n;
  p = build_tree(n);
  for (int i = 1; i <= n; ++i) {
    cin >> x;
    makebase(p + i, x);
  }
  for (int i = p; i >= 1; --i) recount(i);
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> c >> x >> y;
    if (c == 'A')
      cout << find(p + x, p + y - 1, 0) << endl;
    else
      renew(p + x, y);
  }
  return 0;
}
