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;
const int N = 1e5 + 7;
int n, m;
int a[N];
int dp[N << 2][60];
inline void pushup(int rt) {
  int t;
  for (int i = int(0); i <= int(59); ++i) {
    t = (i + dp[rt << 1][i]) % 60;
    dp[rt][i] = dp[rt << 1][i] + dp[rt << 1 | 1][t];
  }
}
void build(int rt, int l, int r) {
  if (l == r) {
    for (int i = int(0); i <= int(59); ++i) {
      if (i % a[l] == 0)
        dp[rt][i] = 2;
      else
        dp[rt][i] = 1;
    }
    return;
  }
  int m = (l + r) >> 1;
  build(rt << 1, l, m);
  build(rt << 1 | 1, m + 1, r);
  pushup(rt);
}
void change(int p, int d, int rt, int l, int r) {
  if (l == r) {
    for (int i = int(0); i <= int(59); ++i) {
      if (i % d == 0)
        dp[rt][i] = 2;
      else
        dp[rt][i] = 1;
    }
    return;
  }
  int m = (l + r) >> 1;
  if (p <= m)
    change(p, d, rt << 1, l, m);
  else
    change(p, d, rt << 1 | 1, m + 1, r);
  pushup(rt);
}
int query(int L, int R, int t, int rt, int l, int r) {
  if (L <= l && r <= R) return dp[rt][t];
  int m = (l + r) >> 1;
  int t1 = 0, t2 = 0;
  if (L <= m) t1 = query(L, R, t, rt << 1, l, m);
  if (R > m) t2 = query(L, R, (t + t1) % 60, rt << 1 | 1, m + 1, r);
  return t1 + t2;
}
int main() {
  scanf("%d", &n);
  for (int i = int(1); i <= int(n); ++i) scanf("%d", a + i);
  build(1, 1, n);
  scanf("%d", &m);
  char s[5];
  int x, y;
  for (int i = int(1); i <= int(m); ++i) {
    scanf("%s%d%d", s, &x, &y);
    if (s[0] == 'C')
      change(x, y, 1, 1, n);
    else
      printf("%d\n", query(x, y - 1, 0, 1, 1, n));
  }
}
