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;
void solve();
void precalc();
clock_t start;
bool doing = true;
int main() {
  int t = 1;
  cout.sync_with_stdio(0);
  cin.tie(0);
  precalc();
  cout.precision(10);
  cout << fixed;
  start = clock();
  int testNum = 1;
  while (t--) {
    solve();
  }
  return 0;
}
void precalc() {}
long long binpow(long long q, long long w, long long mod) {
  if (!w) return 1;
  if (w & 1) return q * binpow(q, w - 1, mod) % mod;
  return binpow(q * q % mod, w / 2, mod);
}
long long gcd(long long q, long long w) {
  while (w) {
    q %= w;
    swap(q, w);
  }
  return q;
}
long long n;
long long a[100500];
const long long mod = 60;
const long long shift = 1 << 17;
struct vertex {
  vector<long long> minTime;
  vertex() { minTime.assign(mod, 0); }
  vertex(long long period) {
    minTime.assign(mod, 0);
    for (long long i = 0; i < mod; ++i) {
      minTime[i] = 1;
      if (i % period == 0) minTime[i] = 2;
    }
  }
};
vertex merge(const vertex& l, const vertex& r) {
  vertex res;
  for (long long i = 0; i < mod; ++i) {
    long long cur = (l.minTime[i] + i) % mod;
    res.minTime[i] = l.minTime[i] + r.minTime[cur];
  }
  return res;
}
vertex tree[2 * shift + 5];
void modify(long long l, long long val) {
  tree[l] = vertex(val);
  l /= 2;
  while (l > 0) {
    tree[l] = merge(tree[2 * l], tree[2 * l + 1]);
    l /= 2;
  }
}
vertex rmq(long long l, long long r) {
  if (l > r) return vertex();
  if (l & 1) {
    return merge(tree[l], rmq(l + 1, r));
  }
  if (!(r & 1)) {
    return merge(rmq(l, r - 1), tree[r]);
  }
  return rmq(l / 2, r / 2);
}
long long ask(long long l, long long r) { return rmq(l, r).minTime[0]; }
void solve() {
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (long long i = 0; i < n; ++i) {
    tree[shift + i] = vertex(a[i]);
  }
  for (long long i = shift - 1; i >= 0; --i) {
    tree[i] = merge(tree[2 * i], tree[2 * i + 1]);
  }
  long long Q;
  cin >> Q;
  for (long long w = 0; w < Q; ++w) {
    char c;
    cin >> c;
    if (c == 'A') {
      long long x, y;
      cin >> x >> y;
      --x;
      --y;
      long long res = ask(x + shift, y - 1 + shift);
      cout << res << "\n";
    } else {
      long long x, y;
      cin >> x >> y;
      --x;
      a[x] = y;
      modify(x + shift, y);
    }
  }
}
