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;
const long long maxN = 1e5 + 7, MOD = 1e9 + 7, inf = 1e15, N = 5000 + 7;
int dp[4 * maxN][4][5][6], d[5][5], res, x, y, a[maxN], n, q, l, r;
int left(int x) { return 2 * x + 1; }
int right(int x) { return 2 * x + 2; }
void build(int node, int st, int en) {
  if (en - st < 2) return;
  int mid = (st + en) >> 1;
  build(left(node), st, mid);
  build(right(node), mid, en);
  for (int x = 0; x < 3; x++) {
    for (int y = 0; y < 4; y++) {
      for (int z = 0; z < 5; z++) {
        int carry = dp[left(node)][x][y][z];
        if (a[mid - 1] == 2) {
          if (!((y + carry) % 2)) carry++;
        }
        if (a[mid - 1] == 3) {
          if (!((carry + x) % 3)) carry++;
        }
        if (a[mid - 1] == 4) {
          if (!((y + carry) % 4)) carry++;
        }
        if (a[mid - 1] == 5) {
          if (!((z + carry) % 5)) carry++;
        }
        if (a[mid - 1] == 6) {
          if (!((d[y & 1][x] + carry) % 6)) carry++;
        }
        carry++;
        dp[node][x][y][z] =
            carry +
            dp[right(node)][(x + carry) % 3][(y + carry) % 4][(z + carry) % 5];
      }
    }
  }
}
void update(int node, int st, int en, int index) {
  if (index < st || en <= index) return;
  if (en - st < 2) return;
  int mid = (st + en) >> 1;
  update(left(node), st, mid, index);
  update(right(node), mid, en, index);
  for (int x = 0; x < 3; x++) {
    for (int y = 0; y < 4; y++) {
      for (int z = 0; z < 5; z++) {
        int carry = dp[left(node)][x][y][z];
        if (a[mid - 1] == 2) {
          if (!((y + carry) % 2)) carry++;
        }
        if (a[mid - 1] == 3) {
          if (!((carry + x) % 3)) carry++;
        }
        if (a[mid - 1] == 4) {
          if (!((y + carry) % 4)) carry++;
        }
        if (a[mid - 1] == 5) {
          if (!((z + carry) % 5)) carry++;
        }
        if (a[mid - 1] == 6) {
          if (!((d[y & 1][x] + carry) % 6)) carry++;
        }
        carry++;
        dp[node][x][y][z] =
            carry +
            dp[right(node)][(x + carry) % 3][(y + carry) % 4][(z + carry) % 5];
      }
    }
  }
}
void get(int node, int st, int en, int l, int r, int &res) {
  if (en <= l || r <= st) return;
  if (st >= l && en <= r) {
    int help = dp[node][res % 3][res % 4][res % 5];
    res += help;
    if (en != r) {
      if (!(res % a[en - 1]))
        res += 2;
      else
        res++;
    }
    return;
  }
  int mid = (st + en) >> 1;
  get(left(node), st, mid, l, r, res);
  get(right(node), mid, en, l, r, res);
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(false);
  for (int i = 0; i < 6; i++) d[i % 2][i % 3] = i;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  build(0, 0, n + 1);
  cin >> q;
  char ch;
  while (q--) {
    cin >> ch;
    if (ch == 'A') {
      cin >> l >> r;
      res = 0;
      l--;
      get(0, 0, n + 1, l, r, res);
      cout << res << "\n";
    } else {
      cin >> x >> y;
      x--;
      a[x] = y;
      update(0, 0, n + 1, x);
    }
  }
  return cout << "\n", 0;
}
