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 maxn = 1e5 + 5;
const int mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
struct node {
  int l, r, time[60];
} tree[maxn << 2];
int a[maxn];
void pushup(int root) {
  for (int i = 0; i < 60; ++i) {
    int tp = tree[root << 1].time[i];
    tree[root].time[i] = tp + tree[root << 1 | 1].time[(i + tp) % 60];
  }
}
void update(int root, int l, int r, int id, int val) {
  if (l == r) {
    for (int i = 0; i < 60; ++i) {
      tree[root].time[i] = 1 + !(i % val);
    }
    return;
  }
  int mid = l + r >> 1;
  if (mid >= id)
    update(root << 1, l, mid, id, val);
  else
    update(root << 1 | 1, mid + 1, r, id, val);
  pushup(root);
}
void build(int root, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 60; ++i) {
      tree[root].time[i] = 1 + !(i % a[l]);
    }
    return;
  }
  int mid = l + r >> 1;
  build(root << 1, l, mid);
  build(root << 1 | 1, mid + 1, r);
  pushup(root);
}
int query(int root, int l, int r, int ql, int qr, int t) {
  if (l == ql && r == qr) {
    return tree[root].time[t];
  }
  int mid = l + r >> 1;
  int tmp = 0;
  if (qr <= mid) {
    return query(root << 1, l, mid, ql, qr, t);
  }
  if (ql > mid) return query(root << 1 | 1, mid + 1, r, ql, qr, t);
  tmp = query(root << 1, l, mid, ql, mid, t);
  t += tmp;
  return tmp + query(root << 1 | 1, mid + 1, r, mid + 1, qr, t % 60);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  build(1, 1, n);
  int q;
  scanf("%d", &q);
  char c;
  int l, r;
  while (q--) {
    scanf(" %c %d%d", &c, &l, &r);
    if (c == 'C') {
      update(1, 1, n, l, r);
    } else {
      printf("%d\n", query(1, 1, n, l, r - 1, 0));
    }
  }
  return 0;
}
