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;
int N;
vector<int> arr;
struct BIT {
  vector<vector<int> > tree;
  void init() { tree = vector<vector<int> >(4 * N, vector<int>(60, 0)); }
  void udt(int idx, int val, int l, int r, int n) {
    if (idx < l || r < idx) return;
    if (l == r) {
      for (int i = 0; i < 60; i++) {
        if (i % val == 0)
          tree[n][i] = 2;
        else
          tree[n][i] = 1;
      }
      return;
    }
    int m = (l + r) >> 1;
    udt(idx, val, l, m, 2 * n);
    udt(idx, val, m + 1, r, 2 * n + 1);
    for (int i = 0; i < 60; i++) {
      tree[n][i] = tree[2 * n][i] + tree[2 * n + 1][(i + tree[2 * n][i]) % 60];
    }
    return;
  }
  int qur(int a, int b, int t, int l, int r, int n) {
    if (b < l || r < a) return 0;
    if (a <= l && r <= b) return tree[n][t % 60];
    int m = (l + r) >> 1;
    int tmp1 = qur(a, b, t, l, m, 2 * n);
    int tmp2 = qur(a, b, t + tmp1, m + 1, r, 2 * n + 1);
    return tmp1 + tmp2;
  }
};
BIT bit;
int Q;
int main() {
  scanf("%d", &N);
  arr.resize(N);
  bit.init();
  for (int i = 0; i < N; i++) {
    scanf("%d", &arr[i]);
    bit.udt(i, arr[i], 0, N - 1, 1);
  }
  scanf("%d", &Q);
  for (int i = 0; i < Q; i++) {
    char t;
    scanf("\n%c", &t);
    if (t == 'A') {
      int x, y;
      scanf("%d %d", &x, &y);
      printf("%d\n", bit.qur(x - 1, y - 2, 0, 0, N - 1, 1));
    } else {
      int x, y;
      scanf("%d %d", &x, &y);
      bit.udt(x - 1, y, 0, N - 1, 1);
    }
  }
}
