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(2)
using namespace std;
void qread(int &x) {
  int neg = 1;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = 10 * x + c - '0', c = getchar();
  x *= neg;
}
const int maxn = 100005;
const int maxm = 65;
int n, q;
int a[maxn];
class node {
 public:
  int val[maxm];
  node() { val[0] = -1; }
  node(int ini) {
    for (int(i) = 0; (i) < 60; i++) {
      if (i % ini == 0)
        val[i] = 2;
      else
        val[i] = 1;
    }
  }
};
namespace Segtree {
node t[maxn * 4];
node merge(node a, node b) {
  if (a.val[0] < 0) return b;
  if (b.val[0] < 0) return a;
  node res;
  for (int(i) = 0; (i) < 60; i++)
    res.val[i] = a.val[i] + b.val[(i + a.val[i]) % 60];
  return res;
}
void build(int v, int l, int r) {
  if (l == r) {
    t[v] = node(a[l]);
    return;
  }
  int mid = (l + r) >> 1;
  build(v << 1, l, mid);
  build(v << 1 | 1, mid + 1, r);
  t[v] = merge(t[v << 1], t[v << 1 | 1]);
}
void update(int v, int l, int r, int k) {
  if (l == r) {
    t[v] = node(a[l]);
    return;
  }
  int mid = (l + r) >> 1;
  if (k <= mid)
    update(v << 1, l, mid, k);
  else
    update(v << 1 | 1, mid + 1, r, k);
  t[v] = merge(t[v << 1], t[v << 1 | 1]);
}
node query(int v, int l, int r, int vl, int vr) {
  if (l >= vl && r <= vr) return t[v];
  int mid = (l + r) >> 1;
  node sl = mid >= vl ? query(v << 1, l, mid, vl, vr) : node();
  node sr = mid < vr ? query(v << 1 | 1, mid + 1, r, vl, vr) : node();
  return merge(sl, sr);
}
}  // namespace Segtree
int main() {
  scanf("%d", &n);
  for (int(i) = 1; (i) <= n; i++) scanf("%d", &a[i]);
  Segtree::build(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    char op[2];
    int x, y;
    scanf("%s%d%d", op, &x, &y);
    if (op[0] == 'A') {
      node ans = Segtree::query(1, 1, n, x, y - 1);
      printf("%d\n", ans.val[0]);
    } else {
      a[x] = y;
      Segtree::update(1, 1, n, x);
    }
  }
  return 0;
}
