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;
int n_split, N, M, a[100009], split[10009], split_dr[10009];
int arb[400009][61];
void refresh(int nod, int mij) {
  for (int i = 0; i < 60; i++) {
    int t = arb[nod << 1][i] + 1 + ((arb[nod << 1][i] + i) % a[mij] == 0);
    arb[nod][i] = t + arb[(nod << 1) + 1][(t + i) % 60];
  }
}
void build(int nod, int st, int dr) {
  if (st == dr) return;
  int mij = (st + dr) >> 1;
  build(nod << 1, st, mij);
  build((nod << 1) + 1, mij + 1, dr);
  refresh(nod, mij);
}
void U(int nod, int st, int dr, int poz) {
  if (st == dr) return;
  int mij = (st + dr) >> 1;
  if (poz <= mij)
    U(nod << 1, st, mij, poz);
  else
    U((nod << 1) + 1, mij + 1, dr, poz);
  refresh(nod, mij);
}
void make_split(int nod, int st, int dr, int x, int y) {
  if (x <= st && dr <= y) {
    split[++n_split] = nod;
    split_dr[n_split] = dr;
    return;
  }
  int mij = (st + dr) >> 1;
  if (x <= mij) make_split(nod << 1, st, mij, x, y);
  if (y > mij) make_split((nod << 1) + 1, mij + 1, dr, x, y);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &a[i]);
  build(1, 1, N);
  scanf("%d\n", &M);
  while (M) {
    M--;
    char tip;
    int x, y;
    scanf("%c %d %d\n", &tip, &x, &y);
    if (tip == 'A') {
      n_split = 0;
      make_split(1, 1, N, x, y);
      int T = 0;
      for (int i = 1; i < n_split; i++) {
        int t2 = T + arb[split[i]][T % 60] + 1 +
                 ((arb[split[i]][T % 60] + T) % a[split_dr[i]] == 0);
        T = t2;
      }
      T += arb[split[n_split]][T % 60];
      if (y == N + 1) {
        if (T % a[N] == 0) T++;
        T++;
      }
      printf("%d\n", T);
    } else {
      a[x] = y;
      U(1, 1, N, x);
    }
  }
  return 0;
}
