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 = 1e5 + 5;
const int MAXM = 60;
struct segtree {
  int node[MAXN << 2][MAXM];
  void pushup(int rt) {
    int lson = rt << 1, rson = rt << 1 | 1;
    for (int i = 0; i < MAXM; ++i) {
      int j = (i + node[lson][i]) % MAXM;
      node[rt][i] = node[lson][i] + node[rson][j];
    }
  }
  void build(int rt, int l, int r) {
    if (l == r) {
      int x;
      scanf("%d", &x);
      for (int i = 0; i < MAXM; ++i) {
        node[rt][i] = i % x ? 1 : 2;
      }
      return;
    }
    int m = (l + r) >> 1;
    build(rt << 1, l, m);
    build(rt << 1 | 1, m + 1, r);
    pushup(rt);
  }
  void update(int rt, int l, int r, int x, int a) {
    if (l == r) {
      for (int i = 0; i < MAXM; ++i) {
        node[rt][i] = i % a ? 1 : 2;
      }
      return;
    }
    int m = (l + r) >> 1;
    x <= m ? update(rt << 1, l, m, x, a) : update(rt << 1 | 1, m + 1, r, x, a);
    pushup(rt);
  }
  int query(int rt, int l, int r, int x, int y, int t) {
    if (x <= l && r <= y) return t + node[rt][t % MAXM];
    int m = (l + r) >> 1;
    if (x <= m) t = query(rt << 1, l, m, x, y, t);
    if (m < y) t = query(rt << 1 | 1, m + 1, r, x, y, t);
    return t;
  }
} tree;
int main() {
  int n;
  while (~scanf("%d", &n)) {
    tree.build(1, 1, n);
    int q;
    scanf("%d", &q);
    while (q--) {
      char op[5];
      int x, y;
      scanf("%s%d%d", op, &x, &y);
      if (*op == 'C')
        tree.update(1, 1, n, x, y);
      else
        printf("%d\n", tree.query(1, 1, n, x, y - 1, 0));
    }
  }
  return 0;
}
