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;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
int seg[61][400001], id[61][400001], ar[100001];
void build(int idx, int s, int e) {
  if (s == e) {
    for (int i = (0); i < (60); ++i) {
      if (i % ar[s] == 0) {
        seg[i][idx] = 2;
        id[i][idx] = (i + 2) % 60;
      } else {
        seg[i][idx] = 1;
        id[i][idx] = (i + 1) % 60;
      }
    }
    return;
  }
  int mid = (s + e) / 2;
  build(idx * 2 + 1, s, mid);
  build(idx * 2 + 2, mid + 1, e);
  for (int i = (0); i < (60); ++i) {
    seg[i][idx] = seg[i][idx * 2 + 1] + seg[id[i][idx * 2 + 1]][idx * 2 + 2];
    id[i][idx] = id[id[i][idx * 2 + 1]][idx * 2 + 2];
  }
}
void insert(int idx, int s, int e, int p) {
  if (s == e) {
    for (int i = (0); i < (60); ++i) {
      if (i % ar[s] == 0) {
        seg[i][idx] = 2;
        id[i][idx] = (i + 2) % 60;
      } else {
        seg[i][idx] = 1;
        id[i][idx] = (i + 1) % 60;
      }
    }
    return;
  }
  int mid = (s + e) / 2;
  if (p <= mid)
    insert(idx * 2 + 1, s, mid, p);
  else
    insert(idx * 2 + 2, mid + 1, e, p);
  for (int i = (0); i < (60); ++i) {
    seg[i][idx] = seg[i][idx * 2 + 1] + seg[id[i][idx * 2 + 1]][idx * 2 + 2];
    id[i][idx] = id[id[i][idx * 2 + 1]][idx * 2 + 2];
  }
}
pair<int, int> query(int c, int idx, int s, int e, int st, int ed) {
  if (s == st && e == ed) return make_pair(id[c][idx], seg[c][idx]);
  int mid = (s + e) / 2;
  if (ed <= mid)
    return query(c, idx * 2 + 1, s, mid, st, ed);
  else if (st > mid)
    return query(c, idx * 2 + 2, mid + 1, e, st, ed);
  else {
    pair<int, int> p1 = query(c, idx * 2 + 1, s, mid, st, mid);
    pair<int, int> p2 = query(p1.first, idx * 2 + 2, mid + 1, e, mid + 1, ed);
    return make_pair(p2.first, p1.second + p2.second);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, b = 0, c, d = 0, e = 0, f = 0, g, h = 0, x, y, z = 0;
  cin >> a;
  for (int i = (1); i < (a + 1); ++i) cin >> ar[i];
  build(0, 1, a);
  cin >> b;
  for (int i = (0); i < (b); ++i) {
    string s;
    cin >> s >> c >> d;
    if (s == "C") {
      ar[c] = d;
      insert(0, 1, a, c);
    } else {
      pair<int, int> pp = query(0, 0, 1, a, c, d - 1);
      cout << pp.second << endl;
    }
  }
}
