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;
int n, sz;
int t[300000][60];
int a[100005];
void build(int l, int r, int x) {
  if (l == r || l > n) return;
  int mid = (l + r) >> 1;
  build(l, mid, x + x);
  build(mid + 1, r, x + x + 1);
  for (int i = 0; i < 60; ++i) {
    int res1 = t[x + x][i];
    int res2 = (i + res1) % 60;
    t[x][i] = (res1 + t[x + x + 1][res2]);
  }
}
void update(int l, int r, int ll, int rr, int x) {
  if (l > r || ll > r || l > rr) return;
  if (l == r) return;
  int mid = (l + r) >> 1;
  update(l, mid, ll, rr, x + x);
  update(mid + 1, r, ll, rr, x + x + 1);
  for (int i = 0; i < 60; ++i) {
    int res1 = t[x + x][i];
    int res2 = (i + res1) % 60;
    t[x][i] = (res1 + t[x + x + 1][res2]);
  }
}
int get(int l, int r, int ll, int rr, int x, int y) {
  if (l >= ll && r <= rr) return y + t[x][y % 60];
  int mid = (l + r) >> 1;
  if (rr <= mid)
    return get(l, mid, ll, rr, x + x, y);
  else if (ll > mid)
    return get(mid + 1, r, ll, rr, x + x + 1, y);
  int res = get(l, mid, ll, rr, x + x, y);
  return get(mid + 1, r, ll, rr, x + x + 1, res);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  sz = 1;
  while (sz < n) sz += sz;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    for (int j = 0; j < 60; ++j)
      if (j % a[i] == 0) {
        t[sz + i - 1][j] = 2;
      } else
        t[sz + i - 1][j] = 1;
  }
  int m;
  build(1, sz, 1);
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    char c;
    int x, y;
    cin >> c >> x >> y;
    if (c == 'A') {
      --y;
      cout << get(1, sz, x, y, 1, 0) << '\n';
    } else {
      a[x] = y;
      for (int j = 0; j < 60; ++j)
        if (j % a[x] == 0) {
          t[sz + x - 1][j] = 2;
        } else
          t[sz + x - 1][j] = 1;
      update(1, sz, x, x, 1);
    }
  }
}
