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 N = 2e5 + 10;
struct node {
  int l, r, y, t[61];
} e[N << 2];
int n, m, a[N];
inline int read() {
  int x = 0;
  char y = '*', z = getchar();
  while (z < '0' || z > '9') y = z, z = getchar();
  while (z >= '0' && z <= '9')
    x = (x << 3) + (x << 1) + (z - '0'), z = getchar();
  return y == '-' ? -x : x;
}
void pushup(int k) {
  for (int i = 0; i < 60; i++) {
    int temp = e[k << 1].t[i];
    int nex = (i + temp) % 60;
    e[k].t[i] = temp + e[k << 1 | 1].t[nex];
  }
}
void build(int l, int r, int k) {
  e[k].l = l;
  e[k].r = r;
  if (l == r) {
    for (int i = 0; i < 60; i++) {
      if (i % a[l] == 0) {
        e[k].t[i] = 2;
      } else {
        e[k].t[i] = 1;
      }
    }
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, k << 1);
  build(mid + 1, r, k << 1 | 1);
  pushup(k);
}
inline void update(int x, int y, int k) {
  if (e[k].l == e[k].r) {
    a[x] = y;
    for (int i = 0; i < 60; i++) {
      if (i % a[x] == 0) {
        e[k].t[i] = 2;
      } else {
        e[k].t[i] = 1;
      }
    }
    return;
  }
  int mid = (e[k].l + e[k].r) >> 1;
  if (x <= mid) {
    update(x, y, k << 1);
  } else {
    update(x, y, k << 1 | 1);
  }
  pushup(k);
}
inline int query(int l, int r, int k, int t) {
  if (l <= e[k].l && e[k].r <= r) {
    return t + e[k].t[t % 60];
  }
  int mid = (e[k].l + e[k].r) >> 1;
  if (l <= mid) {
    t = query(l, r, k << 1, t);
  }
  if (r > mid) {
    t = query(l, r, k << 1 | 1, t);
  }
  return t;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
  }
  build(1, n, 1);
  m = read();
  while (m--) {
    int x, y;
    char q = getchar();
    x = read();
    y = read();
    if (q == 'C') {
      update(x, y, 1);
    }
    if (q == 'A') {
      printf("%d\n", query(x, y - 1, 1, 0));
    }
  }
  return 0;
}
