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;
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
int t[4 * 100005][61], b[100005];
void pushup(int k) {
  for (int i = 0; i < 60; i++)
    t[k][i] = t[(k << 1 | 1)][t[(k << 1)][i] % 60] + t[(k << 1)][i] / 60 * 60;
  return;
}
void build(int k, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 60; i++) {
      if (i % b[l])
        t[k][i] = i + 1;
      else
        t[k][i] = i + 2;
    }
    return;
  }
  build((k << 1), l, ((l + r) / 2));
  build((k << 1 | 1), ((l + r) / 2) + 1, r);
  pushup(k);
  return;
}
int query(int k, int l, int r, int nl, int nr, int v) {
  if (l > nr || r < nl) return v;
  if (nl <= l && nr >= r) {
    return t[k][v % 60] + v / 60 * 60;
  }
  int k1 = query((k << 1), l, ((l + r) / 2), nl, nr, v);
  int k2 = query((k << 1 | 1), ((l + r) / 2) + 1, r, nl, nr, k1);
  return k2;
}
void modify(int k, 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])
        t[k][i] = i + 1;
      else
        t[k][i] = i + 2;
    }
    return;
  }
  modify((k << 1), l, ((l + r) / 2), x, y);
  modify((k << 1 | 1), ((l + r) / 2) + 1, r, x, y);
  pushup(k);
  return;
}
int n;
void init() {
  n = read();
  for (int i = 1; i <= n; i++) b[i] = read();
  build(1, 1, n);
  return;
}
void solve() {
  int q = read();
  while (q--) {
    char c[2];
    cin >> c;
    int x = read(), y = read();
    if (c[0] == 'A')
      printf("%d\n", query(1, 1, n, x, y - 1, 0));
    else
      modify(1, 1, n, x, y);
  }
  return;
}
int main() {
  init();
  solve();
  return 0;
}
