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;
namespace stuff {
const int LCM = 60;
struct Node {
  int take[LCM];
  int end[LCM];
};
const int MOD = 1000000007;
int per[111000];
Node tree[1 << 18];
vector<Node> nodes;
int L, R;
Node combine(const Node& lch, const Node& rch) {
  Node res;
  for (int st = 0; st < LCM; st++) {
    res.end[st] = rch.end[lch.end[st]];
    res.take[st] = lch.take[st] + rch.take[lch.end[st]];
  }
  return res;
}
void build(int at, int l, int r) {
  if (l == r) {
    for (int st = 0; st < LCM; st++) {
      if (st % per[l] == 0) {
        tree[at].take[st] = 2;
        tree[at].end[st] = (st + 2) % LCM;
      } else {
        tree[at].take[st] = 1;
        tree[at].end[st] = (st + 1) % LCM;
      }
    }
  } else {
    const int m = (l + r) >> 1;
    build(at + at, l, m);
    build(at + at + 1, m + 1, r);
    tree[at] = combine(tree[at + at], tree[at + at + 1]);
  }
}
void update(int at, int l, int r, int pos) {
  if (l == r) {
    for (int st = 0; st < LCM; st++) {
      if (st % per[l] == 0) {
        tree[at].take[st] = 2;
        tree[at].end[st] = (st + 2) % LCM;
      } else {
        tree[at].take[st] = 1;
        tree[at].end[st] = (st + 1) % LCM;
      }
    }
  } else {
    const int m = (l + r) >> 1;
    if (pos <= m)
      update(at + at, l, m, pos);
    else
      update(at + at + 1, m + 1, r, pos);
    tree[at] = combine(tree[at + at], tree[at + at + 1]);
  }
}
void query(int at, int l, int r) {
  if (L > r || R < l) return;
  if (L <= l && r <= R)
    nodes.push_back(tree[at]);
  else {
    const int m = (l + r) >> 1;
    query(at + at, l, m);
    query(at + at + 1, m + 1, r);
  }
}
void solve(int test_num) {
  int N;
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> per[i];
  build(1, 1, N);
  int Q;
  cin >> Q;
  while (Q-- > 0) {
    char type;
    cin >> skipws >> type;
    if (type == 'A') {
      cin >> L >> R;
      R--;
      nodes.clear();
      query(1, 1, N);
      Node res = nodes.front();
      for (int i = 1; i < (int)nodes.size(); i++) res = combine(res, nodes[i]);
      cout << res.take[0] << endl;
    } else {
      int pos, val;
      cin >> pos >> val;
      per[pos] = val;
      update(1, 1, N, pos);
    }
  }
}
void solve() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  for (int t = 1; t <= T; t++) solve(t);
}
}  // namespace stuff
int main() {
  stuff::solve();
  return 0;
}
