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 = 100005;
struct tree {
  tree *lchild, *rchild;
  int left, right, time[60], wait[60];
} * root;
int N, Q, xc, yc, ans, pre, a[maxn];
char ch;
void update(tree *p) {
  for (int i = 0; i <= 59; i++)
    p->time[i] = p->rchild->time[p->lchild->time[i]],
    p->wait[i] = p->lchild->wait[i] + p->rchild->wait[p->lchild->time[i]];
}
void maketree(tree *&node, int left, int right) {
  tree *p = new tree;
  p->lchild = 0, p->rchild = 0;
  p->left = left, p->right = right;
  node = p;
  if (left == right) {
    for (int i = 0; i <= 59; i++)
      if (i % a[left] == 0)
        p->time[i] = (i + 2) % 60, p->wait[i] = 1;
      else
        p->time[i] = (i + 1) % 60, p->wait[i] = 0;
  } else {
    maketree(node->lchild, left, (left + right) / 2);
    maketree(node->rchild, (left + right) / 2 + 1, right);
    update(node);
  }
}
void dfsit(tree *p, int s) {
  if (p->left > s) return;
  if (p->right < s) return;
  if (p->left == p->right) {
    for (int i = 0; i <= 59; i++)
      if (i % a[p->left] == 0)
        p->time[i] = (i + 2) % 60, p->wait[i] = 1;
      else
        p->time[i] = (i + 1) % 60, p->wait[i] = 0;
  } else {
    dfsit(p->lchild, s);
    dfsit(p->rchild, s);
    update(p);
  }
}
void dfs(tree *p) {
  int l = p->left, r = p->right;
  if ((l > yc) || (r < xc)) return;
  if ((l == xc) && (r <= yc)) {
    ans += p->wait[pre];
    pre = p->time[pre];
    xc = r + 1;
  } else {
    dfs(p->lchild);
    dfs(p->rchild);
  }
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &a[i]);
  maketree(root, 1, N);
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    while (true) {
      scanf("%c", &ch);
      if ((ch == 'C') || (ch == 'A')) break;
    }
    if (ch == 'C') {
      scanf("%d %d", &xc, &yc);
      a[xc] = yc;
      dfsit(root, xc);
    } else {
      scanf("%d %d", &xc, &yc);
      ans = yc - xc, pre = 0;
      --yc;
      dfs(root);
      printf("%d\n", ans);
    }
  }
  return 0;
}
