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[211111];
int dp[211111][60];
const int C = 80;
void calc(int l, int r) {
  for (int i = r - 1; i >= l; i--) {
    if (i == r - 1) {
      int o = 0;
      for (int j = 0; j < 59; j++) {
        if (o == 0) {
          dp[i][j] = 2;
        } else {
          dp[i][j] = 1;
        }
        o++;
        if (o == a[i]) o = 0;
      }
      if (o == 0) {
        dp[i][59] = 2;
      } else {
        dp[i][59] = 1;
      }
      continue;
    }
    int o = 0;
    for (int j = 0; j < 59; j++) {
      if (o == 0) {
        dp[i][j] = 2 + dp[i + 1][(j + 2) % 60];
      } else {
        dp[i][j] = 1 + dp[i + 1][j + 1];
      }
      o++;
      if (o == a[i]) o = 0;
    }
    if (o == 0) {
      dp[i][59] = 2 + dp[i + 1][1];
    } else {
      dp[i][59] = 1 + dp[i + 1][0];
    }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i <= n; i += C) calc(i, min(i + C, n));
  int t;
  scanf("%d\n", &t);
  while (t--) {
    char type;
    scanf("%c", &type);
    if (type == 'C') {
      int x, v;
      scanf("%d%d\n", &x, &v);
      x--;
      a[x] = v;
      calc(x / C * C, min(x / C * C + C, n));
    } else {
      int x, y;
      scanf("%d%d\n", &x, &y);
      x--;
      y--;
      int t = 0;
      while (x % C != 0 && x < y) {
        if (t % a[x] == 0) t++;
        t++;
        x++;
      }
      while (x < y && min(x + C, n) <= y) {
        int g = dp[x][t % 60];
        t += g;
        x = min(x + C, n);
      }
      while (x < y) {
        if (t % a[x] == 0) t++;
        t++;
        x++;
      }
      printf("%d\n", t);
    }
  }
  return 0;
}
