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>
void read(int &a) {
  a = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    a = (a << 1) + (a << 3) + (c ^ 48);
    c = getchar();
  }
}
int tree[400005][60];
int b[100005];
void build(int a, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 60; i++) {
      if (i % b[l]) {
        tree[a][i] = i + 1;
      } else {
        tree[a][i] = i + 2;
      }
    }
    return;
  }
  int mid = (l + r) >> 1;
  build(a << 1, l, mid);
  build(a << 1 | 1, mid + 1, r);
  for (int i = 0; i < 60; i++) {
    tree[a][i] =
        tree[a << 1 | 1][tree[a << 1][i] % 60] + tree[a << 1][i] / 60 * 60;
  }
  return;
}
int check(int a, int l, int r, int l1, int r1, int t) {
  if (l > r1 || r < l1) {
    return t;
  }
  if (l1 <= l && r <= r1) {
    return tree[a][t % 60] + t / 60 * 60;
  }
  int mid = (l + r) >> 1;
  int k1 = check(a << 1, l, mid, l1, r1, t);
  int k2 = check(a << 1 | 1, mid + 1, r, l1, r1, k1);
  return k2;
}
void gai(int a, int l, int r, int x, int y) {
  if (x < l || r < x) {
    return;
  }
  if (l == x && x == r) {
    b[l] = y;
    for (int i = 0; i < 60; i++) {
      if (i % b[l]) {
        tree[a][i] = i + 1;
      } else {
        tree[a][i] = i + 2;
      }
    }
    return;
  }
  int mid = (l + r) >> 1;
  gai(a << 1, l, mid, x, y);
  gai(a << 1 | 1, mid + 1, r, x, y);
  for (int i = 0; i < 60; i++) {
    tree[a][i] =
        tree[a << 1 | 1][tree[a << 1][i] % 60] + tree[(a << 1)][i] / 60 * 60;
  }
  return;
}
int main() {
  int n;
  read(n);
  for (int i = 1; i <= n; i++) {
    read(b[i]);
  }
  build(1, 1, n);
  int q;
  read(q);
  char qu[3];
  for (int i = 1; i <= q; i++) {
    scanf("%s", qu);
    int x, y;
    read(x), read(y);
    if ((*qu) == 'A') {
      printf("%d\n", check(1, 1, n, x, y - 1, 0));
    } else {
      gai(1, 1, n, x, y);
    }
  }
  return 0;
}
