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>
#pragma GCC optimize("O2")
using namespace std;
const int maxn = 1000 * 100 + 5;
const long long inf = 9223372036854775807;
const long long mod = 60;
const long long lg = 60;
int n, m, seg[lg][4 * maxn], ans, a[4 * maxn], sz = 1, t, sum[lg][4 * maxn];
void build(int x, int lx, int rx) {
  if (rx - lx == 1) {
    for (int i = 0; i < lg; i++) {
      if (a[lx] == 0 || i % a[lx] == 0) {
        seg[i][x] = 2;
        sum[i][x] = 2;
      } else {
        seg[i][x] = 1;
        sum[i][x] = 1;
      }
    }
    return;
  }
  int mid = lx + rx >> 1;
  build(x << 1, lx, mid);
  build(x << 1 | 1, mid, rx);
  for (int i = 0; i < lg; i++) {
    seg[i][x] =
        (seg[i][x << 1] + seg[(i + seg[i][x << 1]) % lg][x << 1 | 1]) % mod;
    sum[i][x] = sum[i][x << 1] + sum[(i + sum[i][x << 1]) % mod][x << 1 | 1];
  }
}
void upd(int i, int v, int x = 1, int lx = 0, int rx = sz) {
  if (rx - lx == 1) {
    a[lx] = v;
    for (int i = 0; i < lg; i++) {
      if (a[lx] == 0 || i % a[lx] == 0) {
        seg[i][x] = 2;
        sum[i][x] = 2;
      } else {
        seg[i][x] = 1;
        sum[i][x] = 1;
      }
    }
    return;
  }
  int mid = lx + rx >> 1;
  if (i < mid) {
    upd(i, v, x << 1, lx, mid);
  } else {
    upd(i, v, x << 1 | 1, mid, rx);
  }
  for (int i = 0; i < lg; i++) {
    seg[i][x] =
        (seg[i][x << 1] + seg[(i + seg[i][x << 1]) % lg][x << 1 | 1]) % mod;
    sum[i][x] = sum[i][x << 1] + sum[(i + sum[i][x << 1]) % mod][x << 1 | 1];
  }
}
void get(int l, int r, int x = 1, int lx = 0, int rx = sz) {
  if (rx <= l || lx >= r) {
    return;
  }
  if (lx >= l && rx <= r) {
    ans += sum[t][x];
    t = (t + sum[t][x]) % 60;
    return;
  }
  int mid = lx + rx >> 1;
  get(l, r, x << 1, lx, mid);
  get(l, r, x << 1 | 1, mid, rx);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  while (sz <= n) sz *= 2;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build(1, 0, sz);
  cin >> m;
  char c;
  int l, r;
  for (int i = 1; i <= m; i++) {
    cin >> c >> l >> r;
    if (c == 'A') {
      ans = 0;
      t = 0;
      get(l, r);
      cout << ans << endl;
    } else {
      upd(l, r);
    }
  }
}
