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>
struct node {
  int ar[60];
  node operator+(const node& rhs) {
    node ret;
    for (int i = 0; i < 60; ++i) {
      int tm = (i + this->ar[i]) % 60;
      ret.ar[i] = this->ar[i] + rhs.ar[tm];
    }
    return ret;
  }
  node() { memset(ar, 0, sizeof(ar)); }
} zero;
const int MAXN = 1 << 17;
node SEG[MAXN * 2];
int ar[MAXN];
void upd(int x) {
  int cur = MAXN + x;
  node& n = SEG[cur];
  for (int t = 0; t < 60; ++t) {
    if (t % ar[x] == 0) {
      n.ar[t] = 2;
    } else {
      n.ar[t] = 1;
    }
  }
  cur /= 2;
  while (cur) {
    SEG[cur] = SEG[(cur * 2)] + SEG[(cur * 2 + 1)];
    cur /= 2;
  }
}
node ask(int left, int right, int cur = 1, int l = 0, int r = MAXN) {
  if (left <= l && r <= right) {
    return SEG[cur];
  } else if (left >= r || l >= right) {
    return zero;
  } else {
    return ask(left, right, (cur * 2), l, ((l + r) / 2)) +
           ask(left, right, (cur * 2 + 1), ((l + r) / 2), r);
  }
}
int N, Q;
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", ar + i);
    upd(i);
  }
  scanf("%d", &Q);
  for (int q = 0; q < Q; ++q) {
    char qtype;
    int x, y;
    scanf(" %c", &qtype);
    if (qtype == 'A') {
      scanf("%d %d", &x, &y);
      node n = ask(x, y);
      int ans = n.ar[0];
      printf("%d\n", ans);
    } else {
      scanf("%d", &x);
      scanf("%d", ar + x);
      upd(x);
    }
  }
  return 0;
}
