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 *tree[400005];
void join(int *L, int *R, int *res) {
  if (R == NULL) {
    for (int i = 0; i < 60; i++) {
      res[i] = L[i];
    }
    return;
  }
  if (L == NULL) {
    for (int i = 0; i < 60; i++) {
      res[i] = R[i];
    }
    return;
  }
  for (int i = 0; i < 60; i++) {
    res[i] = L[i] + R[(i + L[i]) % 60];
  }
}
int *upd(int i, int l, int r, int x, int y) {
  if (x > r || x < l) {
    return tree[i];
  }
  if (l == r) {
    for (int j = 0; j < 60; j++) {
      if (j % y == 0)
        tree[i][j] = 2;
      else
        tree[i][j] = 1;
    }
    return tree[i];
  }
  join(upd(i * 2 + 1, l, (l + r) / 2, x, y),
       upd(i * 2 + 2, (l + r) / 2 + 1, r, x, y), tree[i]);
  return tree[i];
}
int *get(int i, int l, int r, int a, int b) {
  if (l >= a && r <= b) {
    return tree[i];
  }
  if (l > b || r < a) {
    return NULL;
  }
  int *L = get(i * 2 + 1, l, (l + r) / 2, a, b);
  int *R = get(i * 2 + 2, (l + r) / 2 + 1, r, a, b);
  int *ret = new int[60];
  join(L, R, ret);
  if (L != NULL && L != tree[i * 2 + 1]) delete L;
  if (R != NULL && R != tree[i * 2 + 2]) delete R;
  return ret;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < 400005; i++) tree[i] = new int[60];
  int V;
  for (int i = 0; i < n; i++) {
    scanf("%d", &V);
    upd(0, 0, n - 1, i, V);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    char c;
    int x, y;
    scanf(" %c %d %d", &c, &x, &y);
    x--;
    y -= 2;
    if (c == 'A') {
      int *ret = get(0, 0, n - 1, x, y);
      printf("%d\n", ret[0]);
      if (ret != NULL && ret != tree[0]) delete ret;
    } else {
      y += 2;
      upd(0, 0, n - 1, x, y);
    }
  }
}
