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>
#pragma warning(disable : 4996)
int N, Q, now;
int A[100000];
int T[2 * 131072][60];
char inp[5];
void set(int n, int L, int R) {
  int i;
  if (L == R) {
    for (i = 0; i < 60; i++) {
      T[n][i] = (i % A[L]) ? 1 : 2;
    }
    return;
  }
  int M = (L + R) >> 1;
  set(2 * n, L, M);
  set(2 * n + 1, M + 1, R);
  for (i = 0; i < 60; i++) {
    T[n][i] = (i % A[L]) ? T[2 * n][i] : (T[2 * n][(i + 1) % 60] + 1);
    T[n][i] += ((i + T[n][i]) % A[M + 1])
                   ? T[2 * n + 1][(i + T[n][i]) % 60]
                   : (T[2 * n + 1][(i + T[n][i] + 1) % 60] + 1);
  }
}
void u(int n, int L, int R, int p) {
  if (R < p || p < L) return;
  int i;
  if (L == R) {
    for (i = 0; i < 60; i++) {
      T[n][i] = (i % A[L]) ? 1 : 2;
    }
    return;
  }
  int M = (L + R) >> 1;
  u(2 * n, L, M, p);
  u(2 * n + 1, M + 1, R, p);
  for (i = 0; i < 60; i++) {
    T[n][i] = (i % A[L]) ? T[2 * n][i] : (T[2 * n][(i + 1) % 60] + 1);
    T[n][i] += ((i + T[n][i]) % A[M + 1])
                   ? T[2 * n + 1][(i + T[n][i]) % 60]
                   : (T[2 * n + 1][(i + T[n][i] + 1) % 60] + 1);
  }
}
void c(int n, int L, int R, int l, int r) {
  if (R < l || r < L) return;
  int i;
  int M = (L + R) >> 1;
  if (l <= L && R <= r) {
    now += T[n][now % 60];
    return;
  }
  c(2 * n, L, M, l, r);
  c(2 * n + 1, M + 1, R, l, r);
}
int main() {
  int i, j, a, b;
  scanf("%d", &N);
  for (i = 0; i < N; i++) {
    scanf("%d", &A[i]);
  }
  set(1, 0, N - 1);
  scanf("%d", &Q);
  for (i = 0; i < Q; i++) {
    scanf("%s", inp);
    if (*inp == 'C') {
      scanf("%d%d", &a, &b), --a;
      A[a] = b;
      u(1, 0, N - 1, a);
    } else {
      scanf("%d%d", &a, &b), --a, b -= 2;
      now = 0;
      c(1, 0, N - 1, a, b);
      printf("%d\n", now);
    }
  }
  return 0;
}
