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;
struct st {
  long long int a[60];
  st() {}
  st(long long int b) { memset(a, 0, sizeof(a)); }
};
st tree[500011];
long long int a[100011 + 1];
void build(long long int node, long long int l, long long int r) {
  if (l == r) {
    for (int i = 0; i < 60; i++) {
      if (i % a[l] == 0) {
        tree[node].a[i] = 2;
      } else
        tree[node].a[i] = 1;
    }
    return;
  }
  long long int mid = (l + r) / 2;
  build(2 * node, l, mid);
  build(2 * node + 1, mid + 1, r);
  long long int u = 0;
  long long int v = 0;
  for (int i = 0; i < 60; i++) {
    u = tree[2 * node].a[i];
    v = tree[2 * node + 1].a[(u + i) % 60];
    tree[node].a[i] = u + v;
  }
}
st nt(0);
bool func(st x, st y) {
  for (int i = 0; i < 60; i++) {
    if (x.a[i] != y.a[i]) return 0;
  }
  return 1;
}
st query(long long int node, long long int l, long long int r,
         long long int second, long long int e) {
  if (second <= l && r <= e) return tree[node];
  if (second > r || l > r || l > e) return nt;
  long long int mid = (l + r) / 2;
  st x = query(2 * node, l, mid, second, e);
  st y = query(2 * node + 1, mid + 1, r, second, e);
  if (func(x, nt)) return y;
  if (func(y, nt)) return x;
  st z;
  long long int u = 0;
  long long int v = 0;
  for (int i = 0; i < 60; i++) {
    u = x.a[i];
    v = y.a[(u + i) % 60];
    z.a[i] = v + u;
  }
  return z;
}
void update(long long int node, long long int l, long long int r,
            long long int ind) {
  if (l == r) {
    for (int i = 0; i < 60; i++) {
      if (i % a[l] == 0) {
        tree[node].a[i] = 2;
      } else
        tree[node].a[i] = 1;
    }
    return;
  }
  long long int mid = (l + r) / 2;
  if (mid >= ind)
    update(2 * node, l, mid, ind);
  else
    update(2 * node + 1, mid + 1, r, ind);
  long long int u = 0;
  long long int v = 0;
  for (int i = 0; i < 60; i++) {
    u = tree[2 * node].a[i];
    v = tree[2 * node + 1].a[(u + i) % 60];
    tree[node].a[i] = u + v;
  }
}
int main() {
  long long int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build(1, 1, n);
  long long int q;
  cin >> q;
  char c;
  long long int x, y;
  while (q--) {
    cin >> c >> x >> y;
    if (c == 'A') {
      cout << query(1, 1, n, x, y - 1).a[0] << "\n";
    } else {
      a[x] = y;
      update(1, 1, n, x);
    }
  }
}
