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 = 1e9;
const int MOD = 1e9 + 7;
const int N = 100100;
int n, a[N], t[60][4 * N], t1[4 * N], mod[10 * N], q;
void build(int v, int tl, int tr) {
  if (tl == tr) {
    for (int i = 0; i < 60; i++) t[i][v] = 0;
    t1[v] = a[tl];
    return;
  }
  int tm = (tl + tr) / 2;
  build(v * 2, tl, tm);
  build(v * 2 + 1, tm + 1, tr);
  for (int i = 0; i < 60; i++) {
    int x = i + t[i][v * 2];
    int k;
    if (x % t1[v * 2] != 0) {
      x = mod[x + 1];
      k = 1;
    } else {
      x = mod[x + 2];
      k = 2;
    }
    t[i][v] = t[i][v * 2] + t[x][v * 2 + 1] + k;
  }
  t1[v] = t1[v * 2 + 1];
}
void update(int v, int tl, int tr, int pos, int val) {
  if (tl == tr) {
    t1[v] = val;
    return;
  }
  int tm = (tl + tr) / 2;
  if (pos <= tm)
    update(v * 2, tl, tm, pos, val);
  else
    update(v * 2 + 1, tm + 1, tr, pos, val);
  for (int i = 0; i < 60; i++) {
    int x = i + t[i][v * 2];
    int k;
    if (x % t1[v * 2] != 0) {
      x = mod[x + 1];
      k = 1;
    } else {
      x = mod[x + 2];
      k = 2;
    }
    t[i][v] = t[i][v * 2] + t[x][v * 2 + 1] + k;
  }
  t1[v] = t1[v * 2 + 1];
}
int get(int v, int tl, int tr, int l, int r, int x) {
  if (l > r) return 0;
  if (l == tl && r == tr) return t[x][v];
  int tm = (tl + tr) / 2;
  if (l > min(r, tm)) return get(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, x);
  int x1 = get(v * 2, tl, tm, l, min(r, tm), x);
  if (max(tm + 1, l) > r) return x1;
  int k;
  if ((x + x1) % t1[v * 2] != 0)
    k = 1;
  else
    k = 2;
  int x2 = get(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, mod[x + x1 + k]);
  return x1 + x2 + k;
}
int main() {
  scanf("%d\n", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("\n");
  for (int i = 0; i < 3 * N; i++) mod[i] = i % 60;
  build(1, 1, n + 1);
  scanf("%d\n", &q);
  for (int i = 1; i <= q; i++) {
    char ch;
    int x, y;
    scanf("%c %d %d\n", &ch, &x, &y);
    if (ch == 'A') {
      printf("%d\n", get(1, 1, n + 1, x, y, 0));
    } else {
      update(1, 1, n + 1, x, y);
    }
  }
  return 0;
}
