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;
struct Node {
  int val[60];
  Node(int x = 0) {
    for (int i = 0; i < 60; i++) {
      if (x == 0)
        val[i] = 0;
      else if (i % x == 0)
        val[i] = 2;
      else {
        val[i] = 1;
      }
    }
  }
} node[(1 << 18) + 100];
Node merg(Node n1, Node n2) {
  Node rt;
  for (int i = 0; i < 60; i++) {
    rt.val[i] = (n1.val[i] + n2.val[(n1.val[i] + i) % 60]);
  }
  return rt;
}
void setval(int idx, int val) {
  idx += (1 << 17);
  node[idx] = Node(val);
  while (idx / 2) {
    idx /= 2;
    node[idx] = merg(node[idx * 2], node[idx * 2 + 1]);
  }
}
Node getans(int x, int y, int L = 0, int R = (1 << 17) - 1, int idx = 1) {
  if (x > R || y < L) return Node();
  if (x <= L && y >= R) return node[idx];
  int mid = (L + R) / 2;
  return merg(getans(x, y, L, mid, idx << 1),
              getans(x, y, mid + 1, R, idx << 1 | 1));
}
int main() {
  int n;
  scanf("%d", &(n));
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &(x));
    setval(i - 1, x);
  }
  int q;
  scanf("%d", &(q));
  while (q--) {
    char s[3];
    scanf("%s", s);
    int x, y;
    scanf("%d%d", &(x), &(y));
    if (s[0] == 'A') {
      if (x == y) {
        puts("0");
        continue;
      }
      Node ans = getans(x - 1, y - 2);
      printf("%d\n", ans.val[0]);
    } else {
      setval(x - 1, y);
    }
  }
  return 0;
}
