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 a[100005], st[100005 << 2][60], mod60[100005 << 1], now;
bool have[60][7];
void build_tree(int i, int dau, int cuoi) {
  int j;
  if (dau == cuoi) {
    for ((j) = 0; (j) < (60); (j)++)
      if (have[j][a[dau]])
        st[i][j] = 2;
      else
        st[i][j] = 1;
    return;
  }
  int giua = (dau + cuoi) >> 1;
  int i1 = (i << 1), i2 = i1 | 1;
  build_tree(i1, dau, giua);
  build_tree(i2, giua + 1, cuoi);
  for ((j) = 0; (j) < (60); (j)++) {
    int t = st[i1][j];
    int cur = mod60[j + t];
    st[i][j] = t + st[i2][cur];
  }
}
void update(int i, int dau, int cuoi, int pos, int val) {
  int j;
  if (dau == cuoi) {
    for ((j) = 0; (j) < (60); (j)++)
      if (have[j][val])
        st[i][j] = 2;
      else
        st[i][j] = 1;
    return;
  }
  int i1 = i << 1, i2 = i1 | 1;
  int giua = (dau + cuoi) >> 1;
  if (pos <= giua)
    update(i1, dau, giua, pos, val);
  else
    update(i2, giua + 1, cuoi, pos, val);
  for ((j) = 0; (j) < (60); (j)++) {
    int t = st[i1][j];
    int cur = mod60[j + t];
    st[i][j] = t + st[i2][cur];
  }
}
int query(int i, int dau, int cuoi, int l, int r) {
  if (dau == l && cuoi == r) {
    int res = st[i][now];
    now = mod60[now + res];
    return res;
  }
  int i1 = (i << 1), i2 = i1 | 1;
  int giua = (dau + cuoi) >> 1;
  if (r <= giua)
    return query(i1, dau, giua, l, r);
  else if (l > giua)
    return query(i2, giua + 1, cuoi, l, r);
  else
    return query(i1, dau, giua, l, giua) +
           query(i2, giua + 1, cuoi, giua + 1, r);
}
void input() {
  int i, n, nq, x, y;
  char op[5];
  for ((x) = 0; (x) < (60); (x)++)
    for ((y) = (2); (y) <= (6); (y)++) have[x][y] = (x % y == 0);
  for ((i) = 0; (i) < (200000); (i)++) mod60[i] = i % 60;
  scanf("%d", &n);
  for ((i) = (1); (i) <= (n); (i)++) scanf("%d", &a[i]);
  build_tree(1, 1, n);
  scanf("%d", &nq);
  while (nq--) {
    scanf("%s %d %d", &op, &x, &y);
    if (op[0] == 'A') {
      now = 0;
      printf("%d\n", query(1, 1, n, x, y - 1));
    } else {
      a[x] = y;
      update(1, 1, n, x, y);
    }
  }
}
int main() {
  input();
  return 0;
}
