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>
const int inf = 0x3f3f3f3f, Inf = 0x7fffffff;
const long long INF = 0x7fffffffffffffff;
const double eps = 1e-10;
unsigned int seed = 19260817;
const unsigned int _RAND_MAX_ = 4294967295u;
__inline__ __attribute__((always_inline)) unsigned int Rand() {
  return seed = seed * 998244353u + 1000000007u;
}
template <typename _Tp>
_Tp gcd(const _Tp &a, const _Tp &b) {
  return (!b) ? a : gcd(b, a % b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp abs(const _Tp &a) {
  return a > 0 ? a : -a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp max(const _Tp &a, const _Tp &b) {
  return a < b ? b : a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp min(const _Tp &a, const _Tp &b) {
  return a < b ? a : b;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmax(_Tp &a, const _Tp &b) {
  (a < b) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmin(_Tp &a, const _Tp &b) {
  (a > b) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) bool _cmp(const _Tp &a,
                                                    const _Tp &b) {
  return abs(a - b) <= eps;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void read(_Tp &x) {
  register char ch(getchar());
  bool f(false);
  while (ch < 48 || ch > 57) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (ch >= 48 && ch <= 57)
    x = (((x << 2) + x) << 1) + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
__inline__ __attribute__((always_inline)) void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
__inline__ __attribute__((always_inline)) int read_str(char *s) {
  register char ch(getchar());
  while (ch == ' ' || ch == '\r' || ch == '\n') ch = getchar();
  register char *tar = s;
  *tar = ch, ch = getchar();
  while (ch != ' ' && ch != '\r' && ch != '\n' && ch != EOF)
    *(++tar) = ch, ch = getchar();
  return tar - s + 1;
}
const int N = 100005;
int a[N];
struct seg_tr {
  struct Node {
    int ls, rs;
    int t[61];
  } f[N << 1];
  int node_cnt;
  __inline__ __attribute__((always_inline)) void PushUp(const int &x) {
    for (int i = 0; i < 60; ++i) {
      f[x].t[i] = f[f[x].ls].t[i] + f[f[x].rs].t[(i + f[f[x].ls].t[i]) % 60];
    }
  }
  int build(const int &l, const int &r) {
    int cur = ++node_cnt;
    if (l == r) {
      for (int i = 0; i < 60; ++i) {
        if (!(i % a[l])) {
          f[cur].t[i] = 2;
        } else {
          f[cur].t[i] = 1;
        }
      }
      return cur;
    }
    int mid = (l + r) >> 1;
    f[cur].ls = build(l, mid);
    f[cur].rs = build(mid + 1, r);
    PushUp(cur);
    return cur;
  }
  void Update(const int &pos, const int &l, const int &r, const int &cur,
              const int &val) {
    if (l == r) {
      for (int i = 0; i < 60; ++i) {
        if (!(i % val)) {
          f[cur].t[i] = 2;
        } else {
          f[cur].t[i] = 1;
        }
      }
      return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid) Update(pos, l, mid, f[cur].ls, val);
    if (pos > mid) Update(pos, mid + 1, r, f[cur].rs, val);
    PushUp(cur);
  }
  int Query(const int &L, const int &R, const int &l, const int &r,
            const int &cur, const int &_) {
    if (L <= l && r <= R) {
      return f[cur].t[_];
    }
    int mid = (l + r) >> 1;
    if (L <= mid && R > mid) {
      int tmp = Query(L, R, l, mid, f[cur].ls, _);
      return tmp + Query(L, R, mid + 1, r, f[cur].rs, (tmp + _) % 60);
    } else if (L <= mid) {
      return Query(L, R, l, mid, f[cur].ls, _);
    } else {
      return Query(L, R, mid + 1, r, f[cur].rs, _);
    }
  }
} tr;
int main() {
  int n;
  read(n);
  for (int i = 1; i <= n; ++i) read(a[i]);
  int root = tr.build(1, n);
  int _;
  read(_);
  char opt;
  int x, y;
  while (_--) {
    opt = getchar();
    while (opt != 'A' && opt != 'C') opt = getchar();
    read(x, y);
    if (opt == 'C') {
      tr.Update(x, 1, n, root, y);
    } else {
      printf("%d\n", tr.Query(x, y - 1, 1, n, root, 0));
    }
  }
  return 0;
}
