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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int M = 1e5 + 5, INF = 0x3f3f3f3f, mod = 1e9 + 7;
int A[M];
struct node {
  int L, R;
  int dp[60];
  node() { memset(dp, 0, sizeof(dp)); }
} tree[M * 4];
node operator+(node x, node y) {
  node ans;
  for (int j = 0; j < 60; j++) {
    int dis = x.dp[j];
    ans.dp[j] = x.dp[j] + y.dp[(j + dis) % 60];
  }
  ans.L = x.L;
  ans.R = y.R;
  return ans;
}
node zero;
struct seg_tree {
  void up(int p) { tree[p] = tree[p * 2] + tree[p * 2 + 1]; }
  void build(int x, int y, int p) {
    tree[p].L = x;
    tree[p].R = y;
    memset(tree[p].dp, 0, sizeof(tree[p].dp));
    if (x == y) {
      int pre = -1;
      for (int j = 120; j >= 0; j--) {
        int a = j % 2, b = j % 3, c = j % 4, d = j % 5, e = j % 6;
        if (a && b && c && d && e)
          pre = j, tree[p].dp[j % 60] = 1;
        else {
          if (~pre) {
            if (A[x] == 2 && !a)
              tree[p].dp[j % 60] = pre - j + 1;
            else if (A[x] == 3 && !b)
              tree[p].dp[j % 60] = pre - j + 1;
            else if (A[x] == 4 && !c)
              tree[p].dp[j % 60] = pre - j + 1;
            else if (A[x] == 5 && !d)
              tree[p].dp[j % 60] = pre - j + 1;
            else if (A[x] == 6 && !e)
              tree[p].dp[j % 60] = pre - j + 1;
            else {
              tree[p].dp[j % 60] = 1;
              pre = j;
            }
          } else {
            if (A[x] == 2 && !a)
              ;
            else if (A[x] == 3 && !b)
              ;
            else if (A[x] == 4 && !c)
              ;
            else if (A[x] == 5 && !d)
              ;
            else if (A[x] == 6 && !e)
              ;
            else {
              tree[p].dp[j % 60] = 1;
              pre = j;
            }
          }
        }
      }
      return;
    }
    int mid = (x + y) >> 1;
    build(x, mid, p * 2);
    build(mid + 1, y, p * 2 + 1);
    up(p);
  }
  void update(int x, int p) {
    if (tree[p].L == tree[p].R) {
      int pre = -1;
      for (int j = 120; j >= 0; j--) {
        int a = j % 2, b = j % 3, c = j % 4, d = j % 5, e = j % 6;
        if (a && b && c && d && e)
          pre = j;
        else {
          if (~pre) {
            if (A[x] == 2 && !a)
              tree[p].dp[j % 60] = pre - j + 1;
            else if (A[x] == 3 && !b)
              tree[p].dp[j % 60] = pre - j + 1;
            else if (A[x] == 4 && !c)
              tree[p].dp[j % 60] = pre - j + 1;
            else if (A[x] == 5 && !d)
              tree[p].dp[j % 60] = pre - j + 1;
            else if (A[x] == 6 && !e)
              tree[p].dp[j % 60] = pre - j + 1;
            else {
              tree[p].dp[j % 60] = 1;
              pre = j;
            }
          } else {
            if (A[x] == 2 && !a)
              ;
            else if (A[x] == 3 && !b)
              ;
            else if (A[x] == 4 && !c)
              ;
            else if (A[x] == 5 && !d)
              ;
            else if (A[x] == 6 && !e)
              ;
            else {
              tree[p].dp[j % 60] = 1;
              pre = j;
            }
          }
        }
      }
      return;
    }
    int mid = (tree[p].L + tree[p].R) >> 1;
    if (x <= mid)
      update(x, p * 2);
    else
      update(x, p * 2 + 1);
    up(p);
  }
  node query(int x, int y, int p) {
    if (x > y) return zero;
    if (tree[p].L == x && tree[p].R == y) return tree[p];
    int mid = (tree[p].L + tree[p].R) >> 1;
    if (y <= mid)
      return query(x, y, p * 2);
    else if (x > mid)
      return query(x, y, p * 2 + 1);
    else
      return query(x, mid, p * 2) + query(mid + 1, y, p * 2 + 1);
  }
} T;
struct node1 {
  int a, b, c, d, e;
  bool operator<(const node1& x) const {
    if (x.a != a) return a < x.a;
    if (x.b != b) return b < x.b;
    if (x.c != c) return c < x.c;
    if (x.d != d) return d < x.d;
    if (x.e != e) return e < x.e;
    return 0;
  }
};
map<node1, int> mp;
int main() {
  int n;
  scanf("%d", &n);
  for (int j = 1; j <= n; j++) scanf("%d", &A[j]);
  T.build(1, n, 1);
  int q;
  scanf("%d", &q);
  while (q--) {
    char op[2];
    scanf("%s", op);
    if (*op == 'C') {
      int x, y;
      scanf("%d%d", &x, &y);
      A[x] = y;
      T.update(x, 1);
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      printf("%d\n", T.query(x, y - 1, 1).dp[0]);
    }
  }
  return 0;
}
