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 n;
int a[100005];
int q;
char type[5];
vector<int> dp[60][20];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 60; ++i)
    for (int j = 0; j < 20; ++j) dp[i][j].resize(n / (1 << j));
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    for (int j = 0; j < 60; ++j) {
      if (j % a[i] == 0)
        dp[j][0][i] = 2;
      else
        dp[j][0][i] = 1;
    }
  }
  for (int j = 1; j < 20; ++j)
    for (int k = 0; k < 60; ++k)
      for (int i = 0; (i + (1 << j)) <= n; ++i) {
        if (i % (1 << j)) continue;
        int I = i / (1 << j);
        dp[k][j][I] = dp[k][j - 1][2 * I] +
                      dp[(k + dp[k][j - 1][2 * I]) % 60][j - 1][2 * I + 1];
      }
  scanf("%d", &q);
  while (q--) {
    scanf("%s", type);
    int x, y;
    scanf("%d%d", &x, &y);
    if (type[0] == 'A') {
      --x;
      --y;
      int ans = 0;
      while (x < y) {
        for (int i = 19; i >= 0; --i) {
          if (x & ((1 << i) - 1)) continue;
          if (x + (1 << i) > y) continue;
          ans += dp[ans % 60][i][x / (1 << i)];
          x += (1 << i);
          break;
        }
      }
      printf("%d\n", ans);
    } else {
      --x;
      a[x] = y;
      for (int j = 0; j < 60; ++j) {
        if (j % y == 0)
          dp[j][0][x] = 2;
        else
          dp[j][0][x] = 1;
      }
      for (int i = 1; i < 20; ++i) {
        if (x & (1 << (i - 1))) x -= (1 << (i - 1));
        int I = x / (1 << i);
        if ((I + 1) * (1 << i) > n) break;
        for (int k = 0; k < 60; ++k) {
          dp[k][i][I] = dp[k][i - 1][2 * I] +
                        dp[(k + dp[k][i - 1][2 * I]) % 60][i - 1][2 * I + 1];
        }
      }
    }
  }
}
