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 MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
int v[60][4 * MAXN];
int ar[MAXN];
void build(int node, int start, int end) {
  if (start == end) {
    for (int j = 0; j < 60; j++) v[j][node] = 0;
    return;
  }
  int mid = (start + end) >> 1;
  build(2 * node, start, mid);
  build(2 * node + 1, mid + 1, end);
  for (int j = 0; j < 60; j++) {
    int t = v[j][2 * node];
    int mv = j + t;
    mv %= 60;
    for (; mv % ar[mid] == 0; t++, mv = (mv + 1) % 60)
      ;
    t++;
    mv = (mv + 1) % 60;
    t += v[mv][2 * node + 1];
    v[j][node] = t;
  }
}
void update(int node, int start, int end, int qs) {
  if (start > end or qs < start or qs > end) return;
  if (start >= qs and end <= qs) return;
  int mid = (start + end) >> 1;
  update(2 * node, start, mid, qs);
  update(2 * node + 1, mid + 1, end, qs);
  for (int j = 0; j < 60; j++) {
    int t = v[j][2 * node];
    int mv = j + t;
    mv %= 60;
    for (; mv % ar[mid] == 0; t++, mv = (mv + 1) % 60)
      ;
    t++;
    mv = (mv + 1) % 60;
    t += v[mv][2 * node + 1];
    v[j][node] = t;
  }
}
int ans, previ;
void query(int node, int start, int end, int qs, int qe) {
  if (start > end or qe < start or qs > end) return;
  if (start >= qs and end <= qe) {
    if (previ == -1)
      ans = v[0][node];
    else {
      for (; ans % ar[previ] == 0; ans++)
        ;
      ans++;
      ans += v[ans % 60][node];
    }
    previ = end;
    return;
  }
  int mid = (start + end) >> 1;
  query(2 * node, start, mid, qs, qe);
  query(2 * node + 1, mid + 1, end, qs, qe);
}
void reset() {
  ans = 0;
  previ = -1;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &ar[i]);
  int q;
  cin >> q;
  build(1, 1, n + 1);
  while (q--) {
    char ch;
    int x, y;
    scanf(" %c%d%d", &ch, &x, &y);
    if (ch == 'C') {
      ar[x] = y;
      update(1, 1, n + 1, x);
    } else {
      reset();
      query(1, 1, n + 1, x, y);
      printf("%d\n", ans);
    }
  }
  return 0;
}
