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 inf = 0x3f3f3f3f;
typedef struct node2 {
  int d[60 + 1];
} node1;
node1 tree[400001];
int a[100001];
void initialise(int node, int b, int e) {
  if (b > e) {
    return;
  } else if (b == e) {
    for (int(i) = (0); (i) < (60); ++(i)) {
      tree[node].d[i] = 0;
    }
    return;
  } else {
    initialise(2 * node, b, (b + e) / 2);
    initialise(2 * node + 1, (b + e) / 2 + 1, e);
    for (int(i) = (0); (i) < (60); ++(i)) {
      tree[node].d[i] = 0;
    }
    return;
  }
}
void update(int node, int b, int e, int index, int val1) {
  if (b > e) {
    return;
  } else if (b == e && e == index) {
    for (int(i) = (0); (i) < (60); ++(i)) {
      tree[node].d[i] = 1;
      if (i % val1 == 0) {
        tree[node].d[i] += 1;
      }
    }
    return;
  } else {
    if (index <= (b + e) / 2) {
      update(2 * node, b, (b + e) / 2, index, val1);
    } else {
      update(2 * node + 1, (b + e) / 2 + 1, e, index, val1);
    }
    for (int(i) = (0); (i) < (60); ++(i)) {
      tree[node].d[i] = (tree[2 * node].d[i] +
                         tree[2 * node + 1].d[(tree[2 * node].d[i] + i) % 60]);
    }
    return;
  }
}
void query(node1 &res, int node, int b, int e, int i, int j) {
  if (b > e || b > j || e < i) {
    return;
  } else if (i <= b && e <= j) {
    res = tree[node];
    return;
  } else {
    node1 l, r;
    for (int(i) = (0); (i) < (60); ++(i)) {
      l.d[i] = 0;
      r.d[i] = 0;
    }
    query(l, 2 * node, b, (b + e) / 2, i, j);
    query(r, 2 * node + 1, (b + e) / 2 + 1, e, i, j);
    for (int(i) = (0); (i) < (60); ++(i)) {
      res.d[i] = (l.d[i] + r.d[(i + l.d[i]) % 60]);
    }
    return;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  initialise(1, 0, n - 1);
  for (int(i) = (0); (i) < (n); ++(i)) {
    int temp;
    scanf("%d", &temp);
    update(1, 0, n - 1, i, temp);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    char s[2];
    scanf("%s", s);
    if (s[0] == 'A') {
      int first, second;
      scanf("%d%d", &first, &second);
      first--;
      second--;
      second--;
      node1 res;
      for (int(i) = (0); (i) < (30); ++(i)) {
        res.d[i] = 0;
      }
      query(res, 1, 0, n - 1, first, second);
      printf("%d\n", res.d[0]);
    } else {
      int first, second;
      scanf("%d%d", &first, &second);
      first--;
      update(1, 0, n - 1, first, second);
    }
  }
  return 0;
}
