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 n;
int a[2 * 100007];
int id[100007];
struct node {
  int st, en;
  int f[62];
  node() {
    st = en = 0;
    int i;
    for (i = 0; i < 60; i++) {
      f[i] = 0;
    }
  }
  node(int x) {
    st = en = x;
    int i;
    for (i = 0; i < 60; i++) {
      if (a[x] != 1000000 && (i % a[x]) == 0) {
        f[i] = 2;
      } else {
        f[i] = 1;
      }
    }
  }
};
node merge(node p1, node p2) {
  if (p1.st == 0) {
    return p2;
  }
  if (p2.st == 0) {
    return p1;
  }
  int i;
  node ret;
  ret.st = p1.st;
  ret.en = p2.en;
  for (i = 0; i < 60; i++) {
    int tm = i;
    ret.f[i] = p1.f[i] + p2.f[(i + p1.f[i]) % 60];
  }
  return ret;
}
struct segment_tree {
  node tr[5 * 100007];
  void init(int where, int IL, int IR) {
    if (IL > IR) {
      return;
    }
    if (IL == IR) {
      id[IL] = where;
      tr[where] = node(IL);
      return;
    }
    int mid = (IL + IR) / 2;
    init(2 * where, IL, mid);
    init(2 * where + 1, mid + 1, IR);
    tr[where] = merge(tr[2 * where], tr[2 * where + 1]);
  }
  void update(int where, int val) {
    tr[where] = node(tr[where].st);
    where /= 2;
    while (where != 0) {
      tr[where] = merge(tr[2 * where], tr[2 * where + 1]);
      where /= 2;
    }
  }
  node query(int where, int IL, int IR, int CURL, int CURR) {
    if (CURL > CURR) {
      return node(0);
    }
    if (IL > IR) {
      return node(0);
    }
    if (IR < CURL || CURR < IL) {
      return node(0);
    }
    if (CURL <= IL && IR <= CURR) {
      return tr[where];
    }
    int mid = (IL + IR) / 2;
    return merge(query(2 * where, IL, mid, CURL, CURR),
                 query(2 * where + 1, mid + 1, IR, CURL, CURR));
  }
};
segment_tree w;
void input() {
  cin >> n;
  int i;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  a[0] = 1000000;
}
void solve() {
  w.init(1, 1, n);
  int q;
  cin >> q;
  char c;
  int x, y;
  while (q != 0) {
    q--;
    cin >> c;
    cin >> x >> y;
    if (c == 'A') {
      y--;
      cout << w.query(1, 1, n, x, y).f[0] << "\n";
    } else {
      a[x] = y;
      w.update(id[x], y);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}
