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 void rread(T& num) {
  num = 0;
  T f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') num = num * 10 + ch - '0', ch = getchar();
  num *= f;
}
const long long inf = 1e16;
const int maxn = 1e5 + 6, mod = 1e9 + 7;
const long double PI = acos(-1.0);
long long gcd(long long a, long long b) { return (a ? gcd(b % a, a) : b); }
void exgcd(long long a, long long b, long long& d, long long& x, long long& y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    exgcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
int n, m;
int a[maxn];
struct node {
  int r, l;
  int ti[65];
  int tag;
} se[maxn << 2];
void up(int tr) {
  for (int i = 0; i < 60; i++) {
    int d = se[tr << 1].ti[i];
    d = (d + i) % 60;
    se[tr].ti[i] = se[tr << 1].ti[i] + se[tr << 1 | 1].ti[d];
  }
}
void build(int tr, int l, int s) {
  se[tr].r = s;
  se[tr].l = l;
  se[tr].tag = 0;
  int mid = (s + l) / 2;
  if (s == l) {
    for (int i = 0; i < 60; i++) {
      if (i % a[s] == 0)
        se[tr].ti[i] = 2;
      else
        se[tr].ti[i] = 1;
    }
    return;
  }
  build(tr * 2, l, mid);
  build(tr * 2 + 1, mid + 1, s);
  up(tr);
  return;
}
void inse(int tr, int i, int d) {
  if (se[tr].l == se[tr].r) {
    se[tr].tag = d;
    int g = 1;
    for (int i = 0; i < 60; i++)
      if (i % d == 0)
        se[tr].ti[i] = g * 2;
      else
        se[tr].ti[i] = g;
    return;
  }
  int mid = (se[tr].r + se[tr].l) / 2;
  if (i <= mid)
    inse(tr * 2, i, d);
  else
    inse(tr * 2 + 1, i, d);
  up(tr);
  return;
}
int querymax(int tr, int i, int j, int d) {
  if (se[tr].l >= i && se[tr].r <= j) {
    return se[tr].ti[d];
  }
  int mid = (se[tr].r + se[tr].l) >> 1;
  int s1 = 0, s2 = 0;
  if (j <= mid)
    return querymax(tr << 1, i, j, d);
  else if (i > mid)
    return querymax(tr << 1 | 1, i, j, d);
  s1 = querymax(tr << 1, i, j, d);
  d = (d + s1) % 60;
  s2 = querymax(tr << 1 | 1, i, j, d);
  return s1 + s2;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    build(1, 1, n);
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) {
      char s[4];
      int l, r;
      scanf("%s%d%d", s, &l, &r);
      if (s[0] == 'A') {
        r--;
        int ans = querymax(1, l, r, 0);
        printf("%d\n", ans);
      } else {
        inse(1, l, r);
      }
    }
  }
  return 0;
}
