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 agg {
  agg() {}
  agg(int off, int t) {
    off = (t - off % t) % t;
    memset(R, 0, sizeof(R));
    for (int i = off; i < 60; i += t) {
      R[i] = 1;
    }
  }
  int R[60];
};
agg comb(const agg& x, const agg& y) {
  agg r;
  int base = x.R[0] / 60 * 60;
  for (int i = 0; i < 60; i++) {
    int yi = i + x.R[i] - base;
    if (yi >= 60) yi -= 60;
    if (yi < 0) yi += 60;
    r.R[i] = x.R[i] + y.R[yi];
  }
  return r;
}
agg X[2 * (1 << 17)];
agg query(int x, int A, int B, int a, int b) {
  if (a <= A && B <= b) {
    return X[x];
  }
  int M = (A + B) / 2;
  if (b <= M) {
    return query(2 * x + 0, A, M, a, b);
  } else if (M <= a) {
    return query(2 * x + 1, M, B, a, b);
  }
  return comb(query(2 * x + 0, A, M, a, b), query(2 * x + 1, M, B, a, b));
}
int main() {
  ios_base::sync_with_stdio(false);
  int N;
  cin >> N;
  for (int i = 0; i < N; i++) {
    int t;
    cin >> t;
    X[(1 << 17) + i] = agg(i, t);
  }
  for (int i = (1 << 17) - 1; i > 0; i--) {
    X[i] = comb(X[2 * i + 0], X[2 * i + 1]);
  }
  int M;
  cin >> M;
  for (int i = 0; i < M; i++) {
    string cmd;
    int x, y;
    cin >> cmd >> x >> y;
    if (cmd == "C") {
      x--;
      X[(1 << 17) + x] = agg(x, y);
      for (x = ((1 << 17) + x) / 2; x > 0; x /= 2) {
        X[x] = comb(X[2 * x + 0], X[2 * x + 1]);
      }
    } else {
      x--;
      y--;
      int off = (60 - x % 60) % 60;
      cout << y - x + query(1, 0, (1 << 17), x, y).R[off] << '\n';
    }
  }
  return 0;
}
