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 long long UNDEF = -1;
const long long INF = 1e18;
template <typename T>
inline bool chkmax(T& aa, T bb) {
  return aa < bb ? aa = bb, true : false;
}
template <typename T>
inline bool chkmin(T& aa, T bb) {
  return aa > bb ? aa = bb, true : false;
}
static char stdinBuffer[1024];
static char* stdinDataEnd = stdinBuffer + sizeof(stdinBuffer);
static const char* stdinPos = stdinDataEnd;
void readAhead(size_t amount) {
  size_t remaining = stdinDataEnd - stdinPos;
  if (remaining < amount) {
    memmove(stdinBuffer, stdinPos, remaining);
    size_t sz = fread(stdinBuffer + remaining, 1,
                      sizeof(stdinBuffer) - remaining, stdin);
    stdinPos = stdinBuffer;
    stdinDataEnd = stdinBuffer + remaining + sz;
    if (stdinDataEnd != stdinBuffer + sizeof(stdinBuffer)) *stdinDataEnd = 0;
  }
}
int readInt() {
  readAhead(16);
  int x = 0;
  bool neg = false;
  while (*stdinPos == ' ' || *stdinPos == '\n') ++stdinPos;
  if (*stdinPos == '-') {
    ++stdinPos;
    neg = true;
  }
  while (*stdinPos >= '0' && *stdinPos <= '9') {
    x *= 10;
    x += *stdinPos - '0';
    ++stdinPos;
  }
  return neg ? -x : x;
}
char readCh() {
  readAhead(16);
  while (*stdinPos == ' ' || *stdinPos == '\n') ++stdinPos;
  char ans = *stdinPos;
  ++stdinPos;
  return ans;
}
const int mp = 60;
typedef struct S {
  int a[60];
  int len;
} S;
S identity;
S combine(S x, S y) {
  S dest;
  if (x.a[0] == 0) dest = y;
  if (y.a[0] == 0) dest = x;
  for (int i = 0; i < mp; i++) {
    int t = x.a[i];
    dest.a[i] = t + y.a[(i + t) % mp];
  }
  dest.len = x.len + y.len;
  return dest;
}
const long long MAXN = 100002;
S t[2 * MAXN];
int segn;
void build(int n) {
  segn = n;
  for (int i = n - 1; i > 0; --i) {
    t[i] = combine(t[i << 1], t[i << 1 | 1]);
  }
}
void modify(int p, const S& value) {
  int n = segn;
  for (t[p += n] = value; p >>= 1;) t[p] = combine(t[p << 1], t[p << 1 | 1]);
}
S query(int l, int r) {
  int n = segn;
  r++;
  S resl = identity, resr = identity;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      resl = combine(resl, t[l++]);
    }
    if (r & 1) {
      resr = combine(t[--r], resr);
    }
  }
  return combine(resl, resr);
}
S one[7];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n = readInt();
  segn = n + 1;
  for (int x = 2; x <= 6; x++) {
    one[x].len = 1;
    for (int i = 0; i < mp; i++) {
      one[x].a[i] = (i % x == 0) ? 2 : 1;
    }
  }
  for (int i = 0; i < n; i++) {
    int per = readInt();
    t[segn + i] = one[per];
  }
  build(segn);
  S tmp = combine(one[4], one[2]);
  int q = readInt();
  for (int i = 0; i < q; i++) {
    int c = readCh(), x = readInt(), y = readInt();
    if (c == 'A') {
      --x;
      --y;
      S q = query(x, y - 1);
      int ans = q.a[0];
      printf("%d\n", ans);
    } else {
      --x;
      modify(x, one[y]);
    }
  }
}
