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, Q;
int A[100010];
int AI[4 * 100010][121];
void UpdateData(int left[121], int right[121], int v[61]) {
  for (int t = 0; t < 60; t++) v[t] = left[t] + right[(t + left[t]) % 60];
}
void Build(int node, int S, int D) {
  if (S >= D) {
    for (int t = 0; t < 60; t++)
      if (t % A[S] == 0)
        AI[node][t] = 2;
      else
        AI[node][t] = 1;
    return;
  }
  int M = (S + D) / 2;
  Build(2 * node, S, M);
  Build(2 * node + 1, M + 1, D);
  UpdateData(AI[2 * node], AI[2 * node + 1], AI[node]);
}
void Update(int node, int S, int D, int P) {
  if (S >= D) {
    for (int t = 0; t < 60; t++)
      if (t % A[S] == 0)
        AI[node][t] = 2;
      else
        AI[node][t] = 1;
    return;
  }
  int M = (S + D) / 2;
  if (P <= M)
    Update(2 * node, S, M, P);
  else
    Update(2 * node + 1, M + 1, D, P);
  UpdateData(AI[2 * node], AI[2 * node + 1], AI[node]);
}
int ans[61], aux[61];
bool first = 0;
void Query(int node, int S, int D, int L, int R) {
  if (L <= S && D <= R) {
    if (first == 1) {
      UpdateData(ans, AI[node], aux);
      for (int t = 0; t < 60; t++) ans[t] = aux[t];
    } else {
      for (int t = 0; t < 60; t++) ans[t] = AI[node][t];
      first = 1;
    }
    return;
  }
  int M = (S + D) / 2;
  if (L <= M) Query(2 * node, S, M, L, R);
  if (R > M) Query(2 * node + 1, M + 1, D, L, R);
  UpdateData(AI[2 * node], AI[2 * node + 1], AI[node]);
}
int main() {
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> A[i];
  Build(1, 1, N);
  cin >> Q;
  for (int i = 1; i <= Q; i++) {
    char c;
    int x, y;
    cin >> c >> x >> y;
    first = 0;
    if (c == 'A') {
      memset(ans, 0, sizeof(ans));
      Query(1, 1, N, x, y - 1);
      cout << ans[0] << '\n';
    } else {
      A[x] = y;
      Update(1, 1, N, x);
    }
  }
  return 0;
}
