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 N = 1e5;
int n, a[N + 1], segTree[4 * N][60];
void build_tree(int v, int l, int r) {
  if (l == r) {
    for (int Time = 0; Time < 60; Time++) {
      if (Time % a[r] == 0) {
        segTree[v][Time] = 2;
      } else
        segTree[v][Time] = 1;
    }
    return;
  }
  int mid = (l + r) >> 1;
  build_tree(2 * v, l, mid);
  build_tree(2 * v + 1, mid + 1, r);
  for (int Time = 0; Time < 60; Time++) {
    segTree[v][Time] = segTree[2 * v][Time] +
                       segTree[2 * v + 1][(Time + segTree[2 * v][Time]) % 60];
  }
}
void update(int v, int l, int r, int idx, int upd) {
  if (l > r) return;
  if (l == r && r == idx) {
    for (int Time = 0; Time < 60; Time++) {
      if (Time % upd == 0) {
        segTree[v][Time] = 2;
      } else
        segTree[v][Time] = 1;
    }
    return;
  }
  int mid = (l + r) >> 1;
  if (idx <= mid)
    update(2 * v, l, mid, idx, upd);
  else
    update(2 * v + 1, mid + 1, r, idx, upd);
  for (int Time = 0; Time < 60; Time++) {
    segTree[v][Time] = segTree[2 * v][Time] +
                       segTree[2 * v + 1][(Time + segTree[2 * v][Time]) % 60];
  }
}
int query(int v, int tOffset, int l, int r, int start, int end) {
  if (l > r || l > end || r < start) {
    return 0;
  }
  if (l >= start && r <= end) {
    return segTree[v][tOffset];
  }
  int mid = (l + r) >> 1;
  int leftTime = query(2 * v, tOffset, l, mid, start, min(mid, end));
  int rightTime = query(2 * v + 1, (tOffset + leftTime) % 60, mid + 1, r,
                        max(start, mid + 1), end);
  return leftTime + rightTime;
}
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  build_tree(1, 0, n - 1);
  int q;
  scanf("%d", &q);
  for (int ii = 0; ii < q; ii++) {
    char c;
    int x, y;
    scanf(" %c %d %d", &c, &x, &y);
    if (c == 'A') {
      x--;
      y -= 2;
      printf("%d\n", query(1, 0, 0, n - 1, x, y));
    } else {
      x--;
      update(1, 0, n - 1, x, y);
    }
  }
  return 0;
}
