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 MAXN = 300000;
int n, m, x, y, a[MAXN], Left[MAXN], Right[MAXN], cost[MAXN][60];
char c;
void update(int p) {
  for (int i = 0; i < 60; i++) {
    int tmp = cost[p * 2][i];
    cost[p][i] = tmp + cost[p * 2 + 1][(i + tmp) % 60];
  }
}
void build(int p, int l, int r) {
  Left[p] = l;
  Right[p] = r;
  if (l + 1 == r) {
    for (int i = 0; i < 60; i++) {
      cost[p][i] = (i % a[l] != 0) ? 1 : 2;
    }
  } else {
    int mid = (l + r) / 2;
    build(p * 2, l, mid);
    build(p * 2 + 1, mid, r);
    update(p);
  }
}
void modify(int p, int x) {
  if (Left[p] == x && Right[p] == x + 1) {
    for (int i = 0; i < 60; i++) {
      cost[p][i] = (i % a[x] != 0) ? 1 : 2;
    }
    return;
  }
  int mid = (Left[p] + Right[p]) / 2;
  if (x >= mid) {
    modify(p * 2 + 1, x);
  } else {
    modify(p * 2, x);
  }
  update(p);
}
int query(int p, int l, int r, int t) {
  if (Left[p] == l && Right[p] == r) {
    return cost[p][t];
  }
  int mid = (Left[p] + Right[p]) / 2;
  if (r <= mid) {
    return query(p * 2, l, r, t);
  }
  if (l >= mid) {
    return query(p * 2 + 1, l, r, t);
  }
  int tmp = query(p * 2, l, mid, t);
  return tmp + query(p * 2 + 1, mid, r, (t + tmp) % 60);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  build(1, 0, n);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf(" %c %d", &c, &x);
    x--;
    if (c == 'A') {
      scanf("%d", &y);
      y--;
      printf("%d\n", query(1, x, y, 0));
    } else {
      scanf("%d", &a[x]);
      modify(1, x);
    }
  }
  return 0;
}
