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;
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double EPS = 1e-6;
const int INF = 0x3fffffff;
const long long LINF = INF * 1ll * INF;
const double PI = acos(-1.0);
using namespace std;
int a[100005];
struct Node {
  int t[60];
  void init(int period) {
    for (int i = 0; i < 60; i++) {
      if (i % period == 0)
        t[i] = 2;
      else
        t[i] = 1;
    }
  }
} nd[100005 << 2];
class SegTree {
 public:
  int ql, qr, val;
  void build(int l, int r, int rt) {
    if (l == r) {
      nd[rt].init(a[l]);
    } else {
      int mid = (l + r) >> 1;
      build(l, mid, rt << 1);
      build(mid + 1, r, rt << 1 | 1);
      push_Up(l, r, rt);
    }
  }
  void push_Up(int l, int r, int rt) {
    for (int i = 0; i < 60; i++) {
      nd[rt].t[i] = nd[rt << 1].t[i];
      nd[rt].t[i] += nd[rt << 1 | 1].t[(i + nd[rt].t[i]) % 60];
    }
  }
  void update(int l, int r, int rt) {
    if (l == r)
      nd[rt].init(val);
    else {
      int mid = (l + r) >> 1;
      if (ql <= mid)
        update(l, mid, rt << 1);
      else
        update(mid + 1, r, rt << 1 | 1);
      push_Up(l, r, rt);
    }
  }
  int query(int l, int r, int rt, int md) {
    if (ql <= l && r <= qr) {
      if (ql == 3 && qr == 8)
        cout << l << " " << r << " " << md << " " << nd[rt].t[md] << endl;
      return nd[rt].t[md];
    } else {
      int mid = (l + r) >> 1;
      int ans = 0;
      if (ql <= mid) ans += query(l, mid, rt << 1, md);
      md = (md + ans) % 60;
      if (mid < qr) ans += query(mid + 1, r, rt << 1 | 1, md);
      return ans;
    }
  }
} seg;
int main(void) {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int m;
  scanf("%d", &m);
  seg.build(1, n, 1);
  while (m--) {
    char s[2];
    int l, r;
    scanf("%s %d %d", s, &l, &r);
    if (s[0] == 'C') {
      seg.ql = l;
      seg.val = r;
      seg.update(1, n, 1);
    } else {
      seg.ql = l, seg.qr = r - 1;
      printf("%d\n", seg.query(1, n, 1, 0));
    }
  }
  return 0;
}
