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 INF = 0x7f7f7f7f;
const int mo = 1e9 + 7;
int n, q, a[100005];
int L[100005 << 2], R[100005 << 2], sum[100005 << 2][65];
void pushup(int rt) {
  for (int i = 0; i < 60; i++)
    sum[rt][i] = sum[rt << 1][i] + sum[rt << 1 | 1][(i + sum[rt << 1][i]) % 60];
}
void build(int rt, int l, int r) {
  L[rt] = l;
  R[rt] = r;
  int mid = l + r >> 1;
  if (l == r) {
    for (int i = 59; i >= 0; i--) sum[rt][i] = 1 + (i % a[l] == 0);
    return;
  }
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  pushup(rt);
}
void modify(int rt, int ai, int aw) {
  if (L[rt] == R[rt]) {
    a[ai] = aw;
    for (int i = 59; i >= 0; i--) sum[rt][i] = 1 + (i % a[ai] == 0);
    return;
  }
  int mid = L[rt] + R[rt] >> 1;
  if (ai <= mid)
    modify(rt << 1, ai, aw);
  else
    modify(rt << 1 | 1, ai, aw);
  pushup(rt);
}
int query(int rt, int al, int ar, int at) {
  if (L[rt] > ar || R[rt] < al) return 0;
  if (al <= L[rt] && R[rt] <= ar) return sum[rt][at];
  int res = query(rt << 1, al, ar, at);
  res += query(rt << 1 | 1, al, ar, (at + res) % 60);
  return res;
}
signed main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, 1, n);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    char opt[5];
    int x, c, l, r;
    scanf("\n%s", opt);
    if (opt[0] == 'C')
      scanf("%d %d", &x, &c), modify(1, x, c);
    else
      scanf("%d %d", &l, &r), printf("%d\n", query(1, l, r - 1, 0));
  }
  return 0;
}
