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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 1e5 + 100;
const long long mod = 1e9 + 7;
const long double PI = acos((long double)-1);
long long pw(long long a, long long b, long long md = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (a * res) % md;
    }
    a = (a * a) % md;
    b >>= 1;
  }
  return (res);
}
int n;
int a[maxn];
int seg[maxn * 4][70];
int q;
void build(int v = 1, int l = 1, int r = n + 1) {
  if (r - l == 1) {
    for (int i = 0; i < 60; i++) seg[v][i] = 1 + (i % a[l] == 0);
    return;
  }
  int mid = (l + r) / 2;
  build(2 * v, l, mid);
  build(2 * v + 1, mid, r);
  for (int i = 0; i < 60; i++)
    seg[v][i] = seg[2 * v][i] + seg[2 * v + 1][(i + seg[2 * v][i]) % 60];
}
void update(int pos, int x, int v = 1, int l = 1, int r = n + 1) {
  if (r - l == 1) {
    a[l] = x;
    for (int i = 0; i < 60; i++) seg[v][i] = 1 + (i % a[l] == 0);
    return;
  }
  int mid = (l + r) / 2;
  if (pos < mid)
    update(pos, x, 2 * v, l, mid);
  else
    update(pos, x, 2 * v + 1, mid, r);
  for (int i = 0; i < 60; i++)
    seg[v][i] = seg[2 * v][i] + seg[2 * v + 1][(i + seg[2 * v][i]) % 60];
}
int t = 0;
void query(int L, int R, int v = 1, int l = 1, int r = n + 1) {
  if (r <= L or R <= l) {
    return;
  }
  if (L <= l and r <= R) {
    t += seg[v][t % 60];
    return;
  }
  int mid = (l + r) / 2;
  query(L, R, 2 * v, l, mid);
  query(L, R, 2 * v + 1, mid, r);
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build();
  cin >> q;
  while (q--) {
    string t;
    int x, y;
    cin >> t;
    cin >> x >> y;
    if (t == "A") {
      ::t = 0;
      query(x, y);
      cout << ::t << '\n';
    } else {
      update(x, y);
    }
  }
  return (0);
}
