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 inft = 1000000009;
const int MAXN = 1000006;
int SQRT, n;
int t[MAXN];
vector<vector<int> > V;
bool dirty[MAXN][60];
void update(int seg, int i) {
  int czas = i, j = seg * SQRT;
  while (j < min(n, (seg + 1) * SQRT)) {
    if (czas % t[j] != 0) j++;
    czas++;
  }
  V[seg][i] = czas - i;
}
int licz(int a, int b) {
  int czas = 0;
  while (a != b) {
    if (a % SQRT == 0 && a + SQRT <= b) {
      if (dirty[a / SQRT][czas % 60]) {
        dirty[a / SQRT][czas % 60] = false;
        update(a / SQRT, czas % 60);
      }
      czas += V[a / SQRT][czas % 60];
      a += SQRT;
    } else {
      if (czas % t[a] != 0) a++;
      czas++;
    }
  }
  return czas;
}
void solve() {
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) scanf("%d", &t[i]);
  SQRT = 0;
  while (SQRT * SQRT < n) SQRT++;
  V.resize(SQRT + 5, vector<int>(60, 0));
  for (int i = 0; i < (SQRT + 5); ++i)
    for (int j = 0; j < (60); ++j) update(i, j);
  if (0)
    for (int i = 0; i < (SQRT + 5); ++i) {
      printf("seg %d\n", i);
      for (int j = 0; j < (10); ++j) printf("%d ", V[i][j]);
      puts("");
    }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < (q); ++i) {
    char c;
    int a, b;
    scanf(" %c%d%d", &c, &a, &b);
    if (c == 'A') {
      printf("%d\n", licz(a - 1, b - 1));
    } else {
      a--;
      t[a] = b;
      for (int j = 0; j < (60); ++j) dirty[a / SQRT][j] = true;
    }
  }
}
int main() {
  int t = 1;
  for (int i = 0; i < (t); ++i) solve();
  return 0;
}
