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 N = 400100;
void maintain(int* a, int* x, int* y) {
  for (int i = 0; i <= 59; i++) a[i] = x[i] + y[(i + x[i]) % 60];
}
int rec[N];
int val[N][60];
void modify(int k, int l, int r, int x) {
  if (l == r - 1) {
    for (int i = 0; i <= 59; i++) {
      if (i % rec[l] == 0)
        val[k][i] = 2;
      else
        val[k][i] = 1;
    }
    return;
  }
  if (x < ((l + r) >> 1))
    modify((k << 1), l, ((l + r) >> 1), x);
  else
    modify(((k << 1) + 1), ((l + r) >> 1), r, x);
  maintain(val[k], val[(k << 1)], val[((k << 1) + 1)]);
}
int ans[60], tmp[60];
void query(int k, int l, int r, int t, int v) {
  if (v <= l || r <= t) return;
  if (t <= l && r <= v) {
    for (int i = 0; i <= 60; i++) tmp[i] = ans[i];
    maintain(ans, tmp, val[k]);
    return;
  }
  query((k << 1), l, ((l + r) >> 1), t, v);
  query(((k << 1) + 1), ((l + r) >> 1), r, t, v);
}
void Build(int k, int l, int r) {
  if (l == r - 1) {
    for (int i = 0; i <= 59; i++) {
      if (i % rec[l] == 0)
        val[k][i] = 2;
      else
        val[k][i] = 1;
    }
    return;
  }
  Build((k << 1), l, ((l + r) >> 1));
  Build(((k << 1) + 1), ((l + r) >> 1), r);
  maintain(val[k], val[(k << 1)], val[((k << 1) + 1)]);
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", rec + i);
  Build(1, 1, n + 1);
  int q;
  cin >> q;
  while (q--) {
    char c;
    int x, y;
    scanf(" %c%d%d", &c, &x, &y);
    if (c == 'A') {
      for (int i = 0; i <= 59; i++) ans[i] = 0;
      query(1, 1, n + 1, x, y);
      printf("%d\n", ans[0]);
    } else {
      rec[x] = y;
      modify(1, 1, n + 1, x);
    }
  }
  return 0;
}
