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 int N = 1e5 + 10;
const int INF = 1e9 + 19;
const int T = 60;
int n, m;
int a[N];
int tree[N * 4][T];
void read() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  scanf("%d", &m);
}
void upd(int v) {
  for (int i = 0; i < T; i++)
    tree[v][i] =
        tree[v * 2 + 2][(tree[v * 2 + 1][i] + i) % T] + tree[v * 2 + 1][i];
}
void build(int v, int ll, int rr) {
  if (ll + 1 == rr) {
    for (int i = 0; i < T; i++) tree[v][i] = 1 + ((i % a[ll]) == 0);
    return;
  }
  build(v * 2 + 1, ll, (ll + rr) / 2);
  build(v * 2 + 2, (ll + rr) / 2, rr);
  upd(v);
}
int get(int v, int ll, int rr, int l, int r, int time) {
  if (ll >= r || l >= rr) return 0;
  if (l <= ll && rr <= r) {
    return tree[v][time];
  }
  int r1 = get(v * 2 + 1, ll, (ll + rr) / 2, l, r, time);
  int r2 = get(v * 2 + 2, (ll + rr) / 2, rr, l, r, (time + r1) % T);
  return r1 + r2;
}
void apply(int v, int ll, int rr, int pos, int x) {
  if (pos < ll || pos >= rr) return;
  if (ll + 1 == rr) {
    a[pos] = x;
    for (int i = 0; i < T; i++) tree[v][i] = 1 + ((i % a[ll]) == 0);
    return;
  }
  apply(v * 2 + 1, ll, (ll + rr) / 2, pos, x);
  apply(v * 2 + 2, (ll + rr) / 2, rr, pos, x);
  upd(v);
}
void solve() {
  build(0, 0, n);
  for (int i = 0; i < m; i++) {
    char ch;
    int l, r;
    scanf(" %c%d%d", &ch, &l, &r);
    assert(ch == 'A' || ch == 'C');
    if (ch == 'A') {
      l--;
      r--;
      printf("%d\n", get(0, 0, n, l, r, 0));
    } else {
      apply(0, 0, n, l - 1, r);
    }
  }
}
void printAns() {}
void stress() {}
int main() {
  if (1) {
    int k = 1;
    for (int tt = 0; tt < k; tt++) {
      read();
      solve();
      printAns();
    }
  } else {
    stress();
  }
  return 0;
}
