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 MOD = 1000000007;
template <class T>
inline T tmin(T a, T b) {
  return (a < b) ? a : b;
}
template <class T>
inline T tmax(T a, T b) {
  return (a > b) ? a : b;
}
template <class T>
inline void add_max(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline void add_min(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline T tabs(T a) {
  return (a > 0) ? a : -a;
}
template <class T>
T gcd(T a, T b) {
  while (b != 0) {
    T c = a;
    a = b, b = c % b;
  }
  return a;
}
const int N = 100005;
int n, a[N];
struct node {
  vector<int> t;
  node() { t.resize(60); }
};
struct SegmentTree {
  vector<node> seg;
  SegmentTree() {}
  SegmentTree(int _n) {
    int n = 1;
    while (n <= _n) n = n << 1;
    seg.resize(n << 1);
  }
  inline void update_node(int ind) {
    int c1 = ind << 1, c2 = c1 | 1;
    for (int i = (0); i < (60); ++i) {
      seg[ind].t[i] = seg[c1].t[i] + seg[c2].t[(i + seg[c1].t[i]) % 60];
    }
  }
  void build_tree(int ind, int s, int e) {
    if (e - s == 1) {
      seg[ind].t.resize(60);
      for (int i = (0); i < (60); ++i) {
        if (i % a[s])
          seg[ind].t[i] = 1;
        else
          seg[ind].t[i] = 2;
      }
      return;
    }
    int c1 = ind << 1, c2 = c1 | 1, m = (s + e) >> 1;
    build_tree(c1, s, m);
    build_tree(c2, m, e);
    update_node(ind);
  }
  void update(int ind, int s, int e, int idx, int val) {
    if (s > idx || e <= idx) return;
    if (s == idx && e - s == 1) {
      a[s] = val;
      for (int i = (0); i < (60); ++i) {
        if (i % a[s])
          seg[ind].t[i] = 1;
        else
          seg[ind].t[i] = 2;
      }
      return;
    }
    int c1 = ind << 1, c2 = c1 | 1, m = (s + e) >> 1;
    update(c1, s, m, idx, val);
    update(c2, m, e, idx, val);
    update_node(ind);
  }
  vector<int> query(int ind, int s, int e, int ss, int ee) {
    if (s >= ee || e <= ss) return vector<int>(60, 0);
    if (s >= ss && e <= ee) return seg[ind].t;
    int c1 = ind << 1, c2 = c1 | 1, m = (s + e) >> 1;
    vector<int> vl = query(c1, s, m, ss, ee);
    vector<int> vr = query(c2, m, e, ss, ee);
    vector<int> ans(60, 0);
    for (int i = (0); i < (60); ++i) {
      ans[i] = vl[i] + vr[(i + vl[i]) % 60];
    }
    return ans;
  }
};
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  int q, x, y;
  char c;
  cin >> n;
  for (int i = (0); i < (n); ++i) {
    cin >> a[i];
  }
  SegmentTree segtree(n);
  segtree.build_tree(1, 0, n);
  cin >> q;
  for (int i = (0); i < (q); ++i) {
    cin >> c >> x >> y;
    --x;
    if (c == 'A') {
      --y;
      vector<int> ans = segtree.query(1, 0, n, x, y);
      printf("%d\n", ans[0]);
    } else
      segtree.update(1, 0, n, x, y);
  }
  return 0;
}
