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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long MOD = 1000000007;
const long long INF = 1000000000;
template <class T>
T read(T &x) {
  char c;
  int flag = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag *= -1;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = x * 10 + c - '0';
  x *= flag;
  return x;
}
int n;
int lt[100005];
int g[100005];
int dp[100005][65];
int sz = 100;
int a[100005];
char cmd;
int x, y;
int ans;
int st;
int i, j, r, gr;
int main() {
  read(n);
  for (i = 1; i <= n; ++i) read(a[i]);
  for (gr = 1, i = 1; i <= n; i += sz, gr++) {
    lt[gr] = min(n, i + sz - 1);
    g[lt[gr]] = gr;
    for (j = lt[gr] - 1; j >= i; --j) {
      g[j] = gr;
      for (r = 0; r < 60; ++r) {
        if (r % a[j] == 0)
          dp[j][r] = 2 + dp[j + 1][(r + 2) % 60];
        else
          dp[j][r] = 1 + dp[j + 1][(r + 1) % 60];
      }
    }
  }
  int q;
  read(q);
  while (q--) {
    cmd = '#';
    while (!isalpha(cmd)) cmd = getchar();
    read(x);
    read(y);
    if (cmd == 'A') {
      ans = 0;
      while (g[x] != g[y]) {
        ans += dp[x][ans % 60];
        x = lt[g[x]];
        if (ans % a[x] == 0)
          ans += 2;
        else
          ans++;
        x++;
      }
      while (x < y) {
        if (ans % a[x] == 0)
          ans += 2;
        else
          ans++;
        x = x + 1;
      }
      printf("%d\n", ans);
    } else {
      a[x] = y;
      st = (g[x] - 1) * sz + 1;
      if (lt[g[x]] == x) x--;
      for (j = x; j >= st; --j) {
        for (r = 0; r < 60; ++r) {
          if (r % a[j] == 0)
            dp[j][r] = 2 + dp[j + 1][(r + 2) % 60];
          else
            dp[j][r] = 1 + dp[j + 1][(r + 1) % 60];
        }
      }
    }
  }
}
