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 MX = 1 << 17;
int A[MX];
struct node {
  int s[60], t[60];
};
node ini;
struct segtree {
  node t[MX * 2];
  node merge(node x, node y) {
    node ans = {0};
    for (int i = 0; i < 60; i++) {
      ans.s[i] = y.s[x.s[i]];
      ans.t[i] = x.t[i] + y.t[x.s[i]];
    }
    return ans;
  }
  void update(int wi) {
    wi += MX;
    for (int i = 0; i < 60; i++) {
      t[wi].s[i] = (i % A[wi - MX] ? i + 1 : i + 2) % 60;
      t[wi].t[i] = (i % A[wi - MX] ? 1 : 2);
    }
    wi /= 2;
    while (wi) {
      t[wi] = merge(t[wi * 2], t[wi * 2 + 1]);
      wi /= 2;
    }
  }
  node read(int ad, int s, int e, int l, int r) {
    if (e < l || r < s)
      return ini;
    else if (l <= s && e <= r)
      return t[ad];
    else {
      int m = (s + e) / 2;
      return merge(read(ad * 2, s, m, l, r), read(ad * 2 + 1, m + 1, e, l, r));
    }
  }
  node read(int s, int e) { return read(1, 0, MX - 1, s, e); }
} tree;
int main() {
  for (int i = 0; i < 60; i++) ini.s[i] = i;
  for (int i = 0; i < MX; i++) A[i] = 2;
  int N, M;
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d", A + i);
  }
  for (int i = 0; i < MX; i++) tree.update(i);
  scanf("%d", &M);
  for (int t = 1; t <= M; t++) {
    char O[10];
    scanf("%s", O);
    if (O[0] == 'A') {
      int A, B;
      scanf("%d%d", &A, &B);
      B--;
      printf("%d\n", tree.read(A, B).t[0]);
    } else {
      int X, Y;
      scanf("%d%d", &X, &Y);
      A[X] = Y;
      tree.update(X);
    }
  }
  return 0;
}
