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 maxn = 1e5 + 10;
int a[maxn];
struct segtree {
  segtree *lc, *rc;
  int dp[60];
  segtree(int L, int R) {
    lc = rc = nullptr;
    if (L == R) {
      for (int i = 0; i < 60; ++i) {
        dp[i] = 1;
        if (i % a[L] == 0) ++dp[i];
      }
      return;
    }
    lc = new segtree(L, ((L + R) >> 1));
    rc = new segtree(((L + R) >> 1) + 1, R);
    pull();
  }
  void pull() {
    for (int i = 0; i < 60; ++i) {
      int to = (i + lc->dp[i]) % 60;
      dp[i] = lc->dp[i] + rc->dp[to];
    }
  }
  void modify(int L, int R, int p, int v) {
    if (L == R) {
      for (int i = 0; i < 60; ++i) {
        dp[i] = 1;
        if (i % v == 0) ++dp[i];
      }
      return;
    }
    if (p <= ((L + R) >> 1))
      lc->modify(L, ((L + R) >> 1), p, v);
    else
      rc->modify(((L + R) >> 1) + 1, R, p, v);
    pull();
  }
  int query(int L, int R, int l, int r, int idx) {
    if (L > r || l > R) return 0;
    if (L >= l && R <= r) return dp[idx];
    int tl = lc->query(L, ((L + R) >> 1), l, r, idx);
    int to = (idx + tl) % 60;
    return tl + rc->query(((L + R) >> 1) + 1, R, l, r, to);
  }
} * st;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  st = new segtree(1, n);
  int q;
  cin >> q;
  while (q--) {
    char c;
    int x, y;
    cin >> c >> x >> y;
    if (c == 'A')
      cout << st->query(1, n, x, y - 1, 0) << endl;
    else
      st->modify(1, n, x, y);
  }
  return 0;
}
