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;
bool debug = 1;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
int a[100005];
int cal(int t, int i, int e) {
  assert(i <= e);
  if (i == e) return t;
  while (1) {
    while (t % a[i] == 0) {
      t++;
    }
    t++;
    i++;
    if (i == e) break;
  }
  return t;
}
int M, T[270000][60];
char ope[2];
void init() {
  for (int(i) = 0; (i) < (int)(100005); (i)++)
    for (int(j) = 0; (j) < (int)(60); (j)++) T[i][j] = j;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    int id = M + i;
    for (int(j) = 0; (j) < (int)(60); (j)++) T[id][j] = cal(j, i, i + 1);
  }
  for (int i = M - 1; i >= 1; i--) {
    for (int(j) = 0; (j) < (int)(60); (j)++) {
      int tmp = T[i << 1][j];
      T[i][j] = tmp - tmp % 60 + T[i << 1 | 1][tmp % 60];
    }
  }
}
void update(int l, int v) {
  a[l] = v;
  for (int(j) = 0; (j) < (int)(60); (j)++) T[l + M][j] = cal(j, l, l + 1);
  l += M;
  l >>= 1;
  while (l >= 1) {
    for (int(j) = 0; (j) < (int)(60); (j)++) {
      int tmp = T[l << 1][j];
      T[l][j] = tmp - tmp % 60 + T[l << 1 | 1][tmp % 60];
    }
    l >>= 1;
  }
}
int ansL[60], ansR[60], tp[60];
int qy(int l, int r) {
  for (int(j) = 0; (j) < (int)(60); (j)++) {
    ansL[j] = j;
    ansR[j] = j;
  }
  for (l = l + M - 1, r = r + M + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1) {
      for (int(j) = 0; (j) < (int)(60); (j)++) {
        int tmp = ansL[j];
        ansL[j] = tmp - tmp % 60 + T[l ^ 1][tmp % 60];
      }
    }
    if (r & 1) {
      for (int(j) = 0; (j) < (int)(60); (j)++) tp[j] = ansR[j];
      for (int(j) = 0; (j) < (int)(60); (j)++) {
        int tmp = T[r ^ 1][j];
        ansR[j] = tmp - tmp % 60 + tp[tmp % 60];
      }
    }
  }
  int tmp = ansL[0];
  return tmp - tmp % 60 + ansR[tmp % 60];
}
int main() {
  scanf("%d", &n);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", a + i);
  for (M = 1; n + 2 > M; M <<= 1)
    ;
  init();
  scanf("%d", &m);
  int x, y;
  for (int(i) = 0; (i) < (int)(m); (i)++) {
    scanf("%s%d%d", ope, &x, &y);
    if (ope[0] == 'C') {
      update(x, y);
    } else {
      printf("%d\n", qy(x, y - 1));
    }
  }
  return 0;
}
