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;
struct segel {
  vector<int> tm;
  segel(int trp = 0) {
    tm = vector<int>(60, 0);
    if (trp != 0) {
      for (int i = 0; i < (60); ++i) {
        tm[i] = 1;
      }
      for (int i = 0; i < 60; i += trp) tm[i] = 2;
    }
  }
};
void adscar(segel& r, const segel a, const segel& b) {
  for (int i = 0; i < (60); ++i) {
    r.tm[i] = a.tm[i] + b.tm[(i + a.tm[i]) % 60];
  }
}
void ads(segel& r, const segel& a, const segel& b) {
  for (int i = 0; i < (60); ++i) {
    r.tm[i] = a.tm[i] + b.tm[(i + a.tm[i]) % 60];
  }
}
struct segtree {
  int MX;
  const segel neutr = 0;
  vector<segel> mxp;
  vector<int> lft, rgt;
  void mxin(segel& rs, int l, int r, int id = 1) {
    if (r < lft[id] || l > rgt[id]) return;
    if (l <= lft[id] && r >= rgt[id]) {
      adscar(rs, rs, mxp[id]);
      return;
    }
    mxin(rs, l, r, 2 * id);
    mxin(rs, l, r, 2 * id + 1);
  }
  void st(int x, int v, int id = 1) {
    if (x < lft[id] || x > rgt[id]) return;
    if (lft[id] != rgt[id]) {
      st(x, v, 2 * id);
      st(x, v, 2 * id + 1);
      ads(mxp[id], mxp[2 * id], mxp[2 * id + 1]);
    } else
      mxp[id] = v;
  }
  void ini(int n) {
    MX = 1;
    while (MX < n) MX *= 2;
    mxp = vector<segel>(2 * MX, 0);
    lft = rgt = vector<int>(2 * MX);
    for (int i = 0; i < (MX); ++i) lft[i + MX] = rgt[i + MX] = i;
    for (int i = MX - 1; i >= 1; --i) {
      lft[i] = lft[2 * i];
      rgt[i] = rgt[2 * i + 1];
    }
  }
};
segtree mtr;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  mtr.ini(n);
  int tp;
  for (int i = 0; i < (n); ++i) {
    cin >> tp;
    mtr.mxp[mtr.MX + i] = tp;
  }
  for (int i = mtr.MX - 1; i >= 1; --i)
    ads(mtr.mxp[i], mtr.mxp[2 * i], mtr.mxp[2 * i + 1]);
  int q;
  cin >> q;
  char typ;
  int x, y;
  segel tpr;
  for (int z = 0; z < (q); ++z) {
    cin >> typ >> x >> y;
    if (typ == 'A') {
      tpr = 0;
      mtr.mxin(tpr, x - 1, y - 2);
      cout << tpr.tm[0] << "\n";
      continue;
    }
    assert(typ == 'C');
    mtr.st(x - 1, y);
  }
  return 0;
}
