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 mainns {
using std::vector;
template <class RandomAccessIterator, class Function>
class SegmentTree {
 public:
 public:
  SegmentTree(RandomAccessIterator first, RandomAccessIterator last, Function f)
      : first_(first), last_(last), f_(f) {
    valuesHeap_.resize((1 << calcHeight_(first, last)) - 1);
    build_(first, last, 0);
  }
  void update(RandomAccessIterator iter) { update_(iter, 0, first_, last_); }
  typename RandomAccessIterator::value_type getValueOnSegment(
      RandomAccessIterator left, RandomAccessIterator right) const {
    return calcValueOnSegment_(left, right, 0, first_, last_);
  }

 private:
  void build_(RandomAccessIterator left, RandomAccessIterator right,
              size_t ind) {
    if (right - left == 1) {
      valuesHeap_[ind] = *left;
    } else {
      auto center = left + (right - left) / 2;
      build_(left, center, leftChild_(ind));
      build_(center, right, rightChild_(ind));
      valuesHeap_[ind] =
          f_(valuesHeap_[leftChild_(ind)], valuesHeap_[rightChild_(ind)]);
    }
  }
  typename RandomAccessIterator::value_type calcValueOnSegment_(
      RandomAccessIterator segmentLeft, RandomAccessIterator segmentRight,
      size_t nodeInd, RandomAccessIterator nodeLeft,
      RandomAccessIterator nodeRight) const {
    assert(segmentLeft >= nodeLeft);
    assert(segmentRight <= nodeRight);
    if (segmentLeft == nodeLeft && segmentRight == nodeRight)
      return valuesHeap_[nodeInd];
    auto nodeCenter = nodeLeft + (nodeRight - nodeLeft) / 2;
    if (nodeCenter >= segmentRight)
      return calcValueOnSegment_(segmentLeft, segmentRight, leftChild_(nodeInd),
                                 nodeLeft, nodeCenter);
    if (nodeCenter <= segmentLeft)
      return calcValueOnSegment_(segmentLeft, segmentRight,
                                 rightChild_(nodeInd), nodeCenter, nodeRight);
    return f_(calcValueOnSegment_(segmentLeft, nodeCenter, leftChild_(nodeInd),
                                  nodeLeft, nodeCenter),
              calcValueOnSegment_(nodeCenter, segmentRight,
                                  rightChild_(nodeInd), nodeCenter, nodeRight));
  }
  void update_(RandomAccessIterator changed, size_t nodeInd,
               RandomAccessIterator nodeLeft, RandomAccessIterator nodeRight) {
    assert(changed >= nodeLeft);
    assert(changed < nodeRight);
    if (nodeRight - nodeLeft == 1) {
      valuesHeap_[nodeInd] = *changed;
    } else {
      auto nodeCenter = nodeLeft + (nodeRight - nodeLeft) / 2;
      if (changed < nodeCenter)
        update_(changed, leftChild_(nodeInd), nodeLeft, nodeCenter);
      else
        update_(changed, rightChild_(nodeInd), nodeCenter, nodeRight);
      valuesHeap_[nodeInd] = f_(valuesHeap_[leftChild_(nodeInd)],
                                valuesHeap_[rightChild_(nodeInd)]);
    }
  }
  static size_t leftChild_(size_t ind) { return 2 * ind + 1; }
  static size_t rightChild_(size_t ind) { return 2 * ind + 2; }
  int calcHeight_(RandomAccessIterator first, RandomAccessIterator last) const {
    auto length = std::distance(first, last);
    int result = 1;
    size_t maxLen = 1;
    while (length > maxLen) {
      maxLen *= 2;
      result++;
    }
    return result;
  }

 private:
  const RandomAccessIterator first_, last_;
  const Function f_;
  vector<typename RandomAccessIterator::value_type> valuesHeap_;
};
class Solver498D {
 public:
  void run();
  struct TimeToMove {
    int time[60];
  };
  TimeToMove constructFromPeriod(int period) {
    TimeToMove result;
    for (int t = 0; t < 60; ++t) result.time[t] = (t % period == 0) ? 2 : 1;
    return result;
  }
};
void Solver498D::run() {
  int nRoads;
  cin >> nRoads;
  vector<TimeToMove> singleRoadTimes(nRoads);
  for (auto& singleTime : singleRoadTimes) {
    int period;
    cin >> period;
    singleTime = constructFromPeriod(period);
  }
  auto concatFunctor = [](TimeToMove first, TimeToMove second) -> TimeToMove {
    TimeToMove result;
    for (int t = 0; t < 60; ++t) {
      int firstTime = first.time[t];
      result.time[t] = firstTime + second.time[(t + firstTime) % 60];
    }
    return result;
  };
  SegmentTree<decltype(begin(singleRoadTimes)), decltype(concatFunctor)>
      timeTree(begin(singleRoadTimes), end(singleRoadTimes), concatFunctor);
  int nQueries;
  cin >> nQueries;
  for (int i = 0; i < nQueries; ++i) {
    char type;
    int a, b;
    cin >> type >> a >> b;
    if (type == 'A') {
      cout << timeTree
                  .getValueOnSegment(begin(singleRoadTimes) + (a - 1),
                                     begin(singleRoadTimes) + (b - 1))
                  .time[0]
           << endl;
    } else {
      auto iter = begin(singleRoadTimes) + a - 1;
      *iter = constructFromPeriod(b);
      timeTree.update(iter);
    }
  }
}
}  // namespace mainns
using CurrentSolver = mainns::Solver498D;
int main() {
  ios::sync_with_stdio(false);
  CurrentSolver().run();
  return 0;
}
