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 comment(linker, "/STACK:256000000")
using namespace std;
const int maxN = 110000;
const int maxD = 110;
int n;
int a[maxN];
int l[maxN], r[maxN], m;
int t[60][maxN];
int rm[60][maxN];
int id[maxN];
bool dv[60][7];
char s[3];
void calcSeg(int index) {
  int L = l[index];
  int R = r[index];
  for (int i = 0; i < 60; ++i) {
    int value = 0, rem = i;
    int pos = L;
    while (pos != R) {
      if (dv[rem][a[pos]]) {
        ++value;
        ++rem;
        if (rem == 60) {
          rem = 0;
        }
      } else {
        ++value;
        ++rem;
        if (rem == 60) {
          rem = 0;
        }
        ++pos;
      }
    }
    t[i][index] = value;
    rm[i][index] = rem;
  }
}
void precalc() {
  for (int i = 0; i < 60; ++i) {
    for (int j = 1; j <= 6; ++j) {
      dv[i][j] = (i % j == 0);
    }
  }
  int start = 1;
  while (start < n + 1) {
    l[m] = start;
    r[m] = start + maxD;
    if (r[m] >= n + 1) {
      r[m] = n + 1;
    }
    for (int i = l[m]; i < r[m]; ++i) {
      id[i] = m;
    }
    calcSeg(m);
    start = r[m];
    ++m;
  }
  id[n + 1] = m;
}
int getValue(int x, int y) {
  int p = id[x];
  int q = id[y];
  int value = 0;
  int rem = 0;
  int pos = x;
  while (pos != y && id[pos] == p) {
    if (dv[rem][a[pos]]) {
      ++value;
      ++rem;
      if (rem == 60) {
        rem = 0;
      }
    } else {
      ++value;
      ++rem;
      if (rem == 60) {
        rem = 0;
      }
      ++pos;
    }
  }
  if (pos == y) {
    return value;
  }
  while (id[pos] != q) {
    value += t[rem][id[pos]];
    rem = rm[rem][id[pos]];
    pos = r[id[pos]];
  }
  while (pos != y) {
    if (dv[rem][a[pos]]) {
      ++value;
      ++rem;
      if (rem == 60) {
        rem = 0;
      }
    } else {
      ++value;
      ++rem;
      if (rem == 60) {
        rem = 0;
      }
      ++pos;
    }
  }
  return value;
}
void gen() {
  int n = 100000;
  printf("%d\n", n);
  for (int i = 0; i < n; ++i) {
    printf("%d ", (rand() % 5) + 2);
  }
  printf("\n");
  int q = 100000;
  printf("%d\n", q);
  for (int i = 0; i < q; ++i) {
    if (rand() % 10 == 0) {
      int x = 1;
      int y = n + 1;
      printf("A %d %d\n", x, y);
    } else {
      int x = (rand() * rand() + rand()) % n + 1;
      x = ((x % n) + n) % n;
      int y = (rand() % 5) + 2;
      printf("C %d %d\n", x, y);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  precalc();
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%s", &s);
    if (s[0] == 'A') {
      int x, y;
      scanf("%d%d", &x, &y);
      printf("%d\n", getValue(x, y));
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      if (y != a[x]) {
        a[x] = y;
        calcSeg(id[x]);
      }
    }
  }
  return 0;
}
