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>
inline unsigned int getuint() {
  char w = getchar();
  while (w < '0' || '9' < w) w = getchar();
  unsigned int ans = 0;
  for (; '0' <= w && w <= '9'; w = getchar()) ans = ans * 10 + w - '0';
  return ans;
}
inline char getletter() {
  char w = getchar();
  while (w < 'A' || 'Z' < w) w = getchar();
  return w;
}
struct ZKW_SeGment_Tree {
  struct info {
    int t[60];
    info() { memset(t, 0, sizeof(int) * 60); }
    friend inline info operator+(const info &l, const info &r) {
      info ans = l;
      for (int i = 0; i < 60; i++) ans.t[i] += r.t[(i + l.t[i]) % 60];
      return ans;
    }
  } zkw[262144 + 233];
  int shift;
  inline info query(int l, int r) {
    info lans, rans;
    for (l += shift - 1, r += shift + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
      if (~l & 1) lans = lans + zkw[l ^ 1];
      if (r & 1) rans = zkw[r ^ 1] + rans;
    }
    return lans + rans;
  }
  inline void modify(int i, int x) {
    for (int j = 0; j < 60; j++) zkw[i + shift].t[j] = j % x ? 1 : 2;
    for ((i += shift) >>= 1; i; i >>= 1) zkw[i] = zkw[i << 1] + zkw[i << 1 | 1];
  }
  void init(int n, int *a) {
    for (shift = 1; shift <= n + 1; shift <<= 1)
      ;
    for (int i = 1; i <= n; i++)
      for (int j = 0; j < 60; j++) zkw[i + shift].t[j] = j % a[i] ? 1 : 2;
    for (int i = shift - 1; i; i--) zkw[i] = zkw[i << 1] + zkw[i << 1 | 1];
  }
} ZKWSGT;
int main() {
  int n = getuint();
  static int a[100005];
  for (int i = 1; i <= n; i++) a[i] = getuint();
  ZKWSGT.init(n, a);
  for (int m = getuint(); m--;)
    if (getletter() == 'C') {
      int l = getuint(), x = getuint();
      ZKWSGT.modify(l, x);
    } else {
      int l = getuint(), r = getuint();
      printf("%d\n", ZKWSGT.query(l, r - 1).t[0]);
    }
  return 0;
}
