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 Tree {
  int l, r;
  int trans[60], step[60];
} t[100007 << 2];
int n, a[100007];
void Set(int x, int i) {
  for (int j = 0; j < 60; j++) {
    if (j % a[x] == 0)
      t[i].trans[j] = j + 2, t[i].step[j] = 2;
    else
      t[i].trans[j] = j + 1, t[i].step[j] = 1;
    t[i].trans[j] %= 60;
  }
}
void Update(int i) {
  for (int j = 0; j < 60; j++) {
    int tr = t[i << 1].trans[j];
    t[i].trans[j] = t[(i << 1) + 1].trans[tr];
    t[i].step[j] = t[i << 1].step[j] + t[(i << 1) + 1].step[tr];
  }
}
void Build(int l, int r, int i) {
  t[i].l = l, t[i].r = r;
  if (r - l == 1) {
    Set(l, i);
    return;
  }
  int mid = (l + r) >> 1;
  Build(l, mid, i << 1);
  Build(mid, r, (i << 1) + 1);
  Update(i);
}
void Change(int x, int i) {
  if (t[i].r - t[i].l == 1) {
    Set(x, i);
    return;
  }
  int mid = (t[i].l + t[i].r) >> 1;
  if (x < mid)
    Change(x, i << 1);
  else
    Change(x, (i << 1) + 1);
  Update(i);
}
int Query(int l, int r, int i, int& p) {
  if (l >= t[i].r || r <= t[i].l) return 0;
  if (l <= t[i].l && t[i].r <= r) {
    int v = t[i].step[p];
    p = t[i].trans[p];
    return v;
  }
  int v1 = Query(l, r, i << 1, p);
  int v2 = Query(l, r, (i << 1) + 1, p);
  return v1 + v2;
}
int main() {
  char s[10];
  int x, y, q;
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  Build(1, n + 1, 1);
  for (cin >> q; q--;) {
    scanf("%s%d%d", s, &x, &y);
    if (s[0] == 'C') {
      a[x] = y;
      Change(x, 1);
    } else {
      int p = 0;
      printf("%d\n", Query(x, y, 1, p));
    }
  }
  return 0;
}
