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;
int segtree[400005][60];
int vis[400005][60];
int a[100005];
int N, Q;
int iter;
int build(int p, int t, int L, int R) {
  if (vis[p][t % 60]) return segtree[p][t % 60];
  vis[p][t % 60] = 1;
  if (L == R) {
    segtree[p][t % 60] = 1;
    if (t % a[L] == 0) segtree[p][t % 60]++;
    return segtree[p][t % 60];
  }
  int M = (L + R) / 2;
  int left = build(2 * p, t, L, M);
  int right = build(2 * p + 1, t + left, M + 1, R);
  segtree[p][t % 60] = left + right;
  build(2 * p + 1, t, M + 1, R);
  return left + right;
}
int rmq(int p, int t, int L, int R, int S, int T) {
  if (T < L || R < S) return t;
  if (S <= L && R <= T) {
    return t + segtree[p][t % 60];
  }
  int M = (L + R) / 2;
  int left = rmq(2 * p, t, L, M, S, T);
  int right = rmq(2 * p + 1, left, M + 1, R, S, T);
  return right;
}
void update(int p, int L, int R, int S) {
  if (R < S || S < L) return;
  if (R == L && L == S) {
    for (int i = 0; i < 60; ++i) {
      segtree[p][i] = (i % a[L] == 0 ? 2 : 1);
    }
    return;
  }
  int M = (L + R) / 2;
  update(2 * p, L, M, S);
  update(2 * p + 1, M + 1, R, S);
  for (int i = 0; i < 60; ++i) {
    int left = segtree[2 * p][i];
    segtree[p][i] = segtree[2 * p][i] + segtree[2 * p + 1][(i + left) % 60];
  }
}
int main() {
  int x, y;
  char c;
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < 60; ++i) {
    build(1, i, 0, N - 1);
  }
  scanf("%d", &Q);
  iter = 1;
  for (int qq = 0; qq < Q; ++qq) {
    scanf(" %c%d%d", &c, &x, &y);
    if (c == 'C') {
      a[x - 1] = y;
      update(1, 0, N - 1, x - 1);
    } else {
      printf("%d\n", rmq(1, 0, 0, N - 1, x - 1, y - 2));
    }
  }
  return 0;
}
