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;
const int lcm = 60;
class segment_tree {
 private:
  int tree[((1 << 18) + 10)][62];
  int X[100010];
  inline int left(int x) { return (x << 1) + 1; }
  inline int right(int x) { return (x << 1) + 2; }
  int merge(int node, int s, int e, int t) {
    int m, l, r;
    l = left(node);
    r = right(node);
    m = (s + e) >> 1;
    int time = t;
    time += tree[l][time % lcm];
    time += !(time % X[m]);
    ++time;
    time += tree[r][time % lcm];
    return time - t;
  }
  void init(int node, int s, int e, int *A) {
    int i;
    if (s == e) {
      X[s] = A[s];
      for (i = 0; i < lcm; ++i) tree[node][i] = 0;
      return;
    }
    init(left(node), s, (s + e) >> 1, A);
    init(right(node), ((s + e) >> 1) + 1, e, A);
    for (i = 0; i < lcm; ++i) {
      tree[node][i] = merge(node, s, e, i);
    }
  }
  int query(int node, int s, int e, int i, int j, int time) {
    if (s > j || e < i) return time;
    if (s >= i && e <= j) {
      return time + tree[node][time % lcm];
    }
    int l = left(node), r = right(node), m = (s + e) >> 1;
    time = query(l, s, m, i, j, time);
    if (m >= j) return time;
    if (m >= i) {
      time += !(time % X[m]);
      ++time;
    }
    return query(r, m + 1, e, i, j, time);
  }
  void update(int node, int s, int e, int i) {
    if (s > i || e < i) return;
    if (s == i && e == i) return;
    update(left(node), s, (s + e) >> 1, i);
    update(right(node), ((s + e) >> 1) + 1, e, i);
    int k;
    for (k = 0; k < lcm; ++k) {
      tree[node][k] = merge(node, s, e, k);
    }
  }
  int N;

 public:
  void init(int n, int *A) {
    N = n;
    init(0, 0, N - 1, A);
  }
  int query(int i, int j) { return query(0, 0, N - 1, i, j, 0); }
  void update(int i, int val) {
    X[i] = val;
    update(0, 0, N - 1, i);
  }
};
int N, Q;
int A[100010];
segment_tree T;
int main() {
  ios_base::sync_with_stdio(0);
  int i, x, y;
  char c;
  cin >> N;
  for (i = 0; i < N; ++i) cin >> A[i];
  A[N++] = 6;
  T.init(N, A);
  cin >> Q;
  while (Q--) {
    cin >> c >> x >> y;
    if (c ^ 'C') {
      cout << T.query(x - 1, y - 1) << '\n';
      continue;
    }
    --x;
    A[x] = y;
    T.update(x, y);
  }
  return 0;
}
