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;
long long poww(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * poww(a * a % md, b / 2, md) % md
                      : poww(a * a % md, b / 2, md) % md));
}
const long long MAXN = 1e6 + 10;
const long long INF = 8e18;
const long long MOD = 60;
int sg[MOD][MAXN], n, A[MAXN], q;
void Build(int v, int L, int R) {
  if (R < L) return;
  if (R == L) {
    for (int i = 0; i < MOD; i++) {
      if (i % A[L] == 0)
        sg[i][v] = 2;
      else
        sg[i][v] = 1;
    }
    return;
  }
  int mid = (L + R) >> 1;
  Build(v << 1, L, mid);
  Build(v << 1 | 1, mid + 1, R);
  for (int i = 0; i < MOD; i++)
    sg[i][v] = sg[i][v << 1] + sg[(i + sg[i][v << 1]) % MOD][v << 1 | 1];
}
void Update(int v, int L, int R, int ind) {
  if (R < L) return;
  if (R == L) {
    for (int i = 0; i < MOD; i++) {
      if (i % A[L] == 0)
        sg[i][v] = 2;
      else
        sg[i][v] = 1;
    }
    return;
  }
  int mid = (L + R) >> 1;
  if (ind <= mid)
    Update(v << 1, L, mid, ind);
  else
    Update(v << 1 | 1, mid + 1, R, ind);
  for (int i = 0; i < 60; i++)
    sg[i][v] = sg[i][v << 1] + sg[(i + sg[i][v << 1]) % MOD][v << 1 | 1];
}
int Query(int v, int L, int R, int ql, int qr, int mod = 0) {
  if (R < L || R < ql || qr < L) return 0;
  if (ql <= L && qr >= R) return sg[mod][v];
  int mid = (L + R) >> 1;
  int ans = Query(v << 1, L, mid, ql, qr, mod);
  ans += Query(v << 1 | 1, mid + 1, R, ql, qr, (mod + ans) % MOD);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> A[i];
  Build(1, 1, n);
  cin >> q;
  while (q--) {
    char c;
    int x, y;
    cin >> c >> x >> y;
    if (c == 'C') {
      A[x] = y;
      Update(1, 1, n, x);
    } else
      cout << Query(1, 1, n, x, y - 1) << '\n';
  }
  return 0;
}
