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 GAP = 131072, LCM = 60;
int n, tree[GAP << 1][LCM], right[GAP << 1], numQuery;
void init(int pos, int num) {
  for (int i = 0; i < LCM; i++) tree[pos][i] = i % num == 0 ? 2 : 1;
}
void input() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int t;
    scanf("%d", &t);
    init(GAP + i, t);
  }
  scanf("%d", &numQuery);
}
void combine(int num) {
  int l = num * 2, r = num * 2 + 1;
  for (int i = 0; i < LCM; i++)
    tree[num][i] = tree[l][i] + tree[r][(i + tree[l][i]) % LCM];
}
void initTree() {
  for (int i = GAP; i < GAP * 2; i++) right[i] = i - GAP;
  for (int i = GAP - 1; i > 0; i--) {
    combine(i);
    right[i] = right[i * 2 + 1];
  }
}
int query(int f, int s) {
  int current = 0, now = f + GAP;
  while (now <= s + GAP) {
    if ((now & 1) == 0 && right[now >> 1] <= s) {
      now >>= 1;
    } else {
      current += tree[now][current % LCM];
      now = right[now] + GAP + 1;
    }
  }
  return current;
}
void solve() {
  for (int i = 0; i < numQuery; i++) {
    int x, y;
    char type[2];
    scanf("%s%d%d", type, &x, &y);
    if (type[0] == 'C') {
      x--;
      init(GAP + x, y);
      for (int now = ((GAP + x) >> 1); now > 0; now >>= 1) combine(now);
    } else if (type[0] == 'A') {
      printf("%d\n", query(x - 1, y - 2));
    }
  }
}
int main() {
  input();
  initTree();
  solve();
  return 0;
}
