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 inf = (1 << 30) - 1, mod = 1000000007;
const int N = 100005;
int n, m, a[N];
char ch;
struct data {
  int t[60], c[60], last;
  data() {
    memset(t, 0, sizeof(t));
    memset(c, 0, sizeof(c));
  }
  data(int x) {
    memset(c, 0, sizeof(c));
    for (int i = 0; i < 60; i++) t[i] = i;
    last = x;
  }
  data const operator+(const data &d) const {
    data res = data();
    res.last = d.last;
    for (int i = 0; i < 60; ++i) {
      int nxt = t[i] + 1 + (t[i] % last == 0);
      res.c[i] = c[i];
      if (nxt >= 60) {
        nxt -= 60;
        res.c[i]++;
      }
      res.c[i] += d.c[nxt];
      res.t[i] = d.t[nxt];
    }
    return res;
  }
};
data s[4 * N];
void build(int v, int l, int r) {
  if (l == r)
    s[v] = data(a[l]);
  else {
    build(v + v, l, ((l + r) >> 1));
    build(v + v + 1, ((l + r) >> 1) + 1, r);
    s[v] = s[v + v] + s[v + v + 1];
  }
}
void update(int v, int l, int r, int pos, int x) {
  if (l == r)
    s[v] = data(x), a[l] = x;
  else {
    if (pos <= ((l + r) >> 1))
      update(v + v, l, ((l + r) >> 1), pos, x);
    else
      update(v + v + 1, ((l + r) >> 1) + 1, r, pos, x);
    s[v] = s[v + v] + s[v + v + 1];
  }
}
void step(pair<int, int> &x, int last) {
  x.second += 1 + (x.second % last == 0);
  if (x.second >= 60) {
    x.first++;
    x.second -= 60;
  }
}
int get(int v, int l, int r, int L, int R, pair<int, int> &x) {
  if (r < L || R < l) return 0;
  if (L <= l && r <= R) {
    x = make_pair(x.first + s[v].c[x.second], s[v].t[x.second]);
    return 0;
  }
  if (R <= ((l + r) >> 1)) return get(v + v, l, ((l + r) >> 1), L, R, x);
  if (((l + r) >> 1) < L) return get(v + v + 1, ((l + r) >> 1) + 1, r, L, R, x);
  get(v + v, l, ((l + r) >> 1), L, R, x);
  step(x, a[((l + r) >> 1)]);
  return get(v + v + 1, ((l + r) >> 1) + 1, r, L, R, x);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  build(1, 1, n);
  scanf("%d\n", &m);
  for (int i = 0, x, y; i < m; i++) {
    scanf("%c %d %d\n", &ch, &x, &y);
    if (ch == 'C')
      update(1, 1, n, x, y);
    else {
      pair<int, int> res = make_pair(0, 0);
      y--;
      get(1, 1, n, x, y, res);
      step(res, a[y]);
      printf("%d\n", res.first * 60 + res.second);
    }
  }
  return 0;
}
