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 int N = 1e5 + 5;
int out[N], n;
struct node {
  int val[61];
  node() {}
} tree[N << 2];
node merge(node a, node b, int mod) {
  node ret;
  for (int i = 0; i <= 60; i++) {
    int ans = a.val[i];
    int out_time = (a.val[i] + i) % 60;
    ans += b.val[out_time];
    ret.val[i] = ans;
  }
  return ret;
}
int ans[105][105];
void init() {
  for (int i = 0; i <= 60; i++) {
    for (int j = 2; j <= 60; j++) {
      int time = i;
      while (time % j == 0) time++;
      time++;
      ans[i][j] = (time - i);
    }
  }
}
int solve(int s, int e) {
  int cur = 0;
  while (s <= e) {
    while (cur % out[s] == 0) cur++;
    cur++;
    s++;
  }
  return cur;
}
void build(int node, int s, int e) {
  if (s == e) {
    for (int i = 0; i <= 60; i++) {
      tree[node].val[i] = ans[i][out[s]];
    }
    return;
  }
  int md = (s + e) >> 1;
  build(node << 1, s, md);
  build(node << 1 | 1, md + 1, e);
  tree[node] = merge(tree[node << 1], tree[node << 1 | 1], out[md + 1]);
}
void upd(int node, int s, int e, int idx, int v) {
  if (s == e) {
    out[s] = v;
    for (int i = 0; i <= 60; i++) {
      tree[node].val[i] = ans[i][out[s]];
    }
    return;
  }
  int md = (s + e) >> 1;
  if (idx <= md)
    upd(node << 1, s, md, idx, v);
  else
    upd(node << 1 | 1, md + 1, e, idx, v);
  tree[node] = merge(tree[node << 1], tree[node << 1 | 1], out[md + 1]);
}
node qry(int node, int s, int e, int l, int r) {
  if (l <= s && e <= r) return tree[node];
  int md = (s + e) >> 1;
  auto ret = tree[node];
  if (md >= l && md + 1 <= r)
    ret = merge(qry(node << 1, s, md, l, r),
                qry(node << 1 | 1, md + 1, e, l, r), out[md + 1]);
  else if (md >= l)
    ret = qry(node << 1, s, md, l, r);
  else
    ret = qry(node << 1 | 1, md + 1, e, l, r);
  return ret;
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", out + i);
  build(1, 0, n - 1);
  int q, x, y;
  char c;
  scanf("%d", &q);
  while (q--) {
    scanf(" %c%d%d", &c, &x, &y);
    if (c == 'C')
      upd(1, 0, n - 1, x - 1, y);
    else
      printf("%d\n", qry(1, 0, n - 1, x - 1, y - 2).val[0]);
  }
  return 0;
}
