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;
struct node {
  int lef, rig, val[65];
} nop, nop2, nop3;
int n, tim[120000];
inline node uni(const node &x, const node &y) {
  node ntm;
  for (int i = 0; i < 60; i++) {
    ntm.val[i] = x.val[i] + y.val[(i + x.val[i]) % 60];
  }
  return ntm;
}
struct segement {
  node a[520000];
  inline void update(int x) {
    for (int i = 0; i < 60; i++) {
      a[x].val[i] =
          a[x * 2].val[i] + a[x * 2 + 1].val[(i + a[x * 2].val[i]) % 60];
    }
  }
  inline void build(int x, int le, int ri) {
    a[x].lef = le;
    a[x].rig = ri;
    if (le == ri) {
      for (int i = 0; i < 60; i++) {
        if (i % tim[le] == 0)
          a[x].val[i] = 2;
        else
          a[x].val[i] = 1;
      }
      return;
    }
    build(x * 2, le, (le + ri) / 2);
    build(x * 2 + 1, (le + ri) / 2 + 1, ri);
    update(x);
  }
  inline void edit(int x, int pos, int va) {
    if (a[x].lef == a[x].rig) {
      for (int i = 0; i < 60; i++) {
        if (i % tim[a[x].lef] == 0)
          a[x].val[i] = 2;
        else
          a[x].val[i] = 1;
      }
      return;
    }
    if (pos <= a[x * 2].rig)
      edit(x * 2, pos, va);
    else
      edit(x * 2 + 1, pos, va);
    update(x);
  }
  inline node query(int x, int le, int ri) {
    if (le <= a[x].lef && ri >= a[x].rig) {
      return a[x];
    }
    node np1;
    if (le <= a[x * 2].rig && ri >= a[x * 2 + 1].lef) {
      np1 = uni(query(x * 2, le, ri), query(x * 2 + 1, le, ri));
    } else if (le <= a[x * 2].rig) {
      np1 = query(x * 2, le, ri);
    } else {
      np1 = query(x * 2 + 1, le, ri);
    }
    return np1;
  }
  inline int query(int le, int ri) { return query(1, le, ri).val[0]; }
} sgt;
int m, x1, x2, x3;
char chx[20];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &tim[i]);
  }
  sgt.build(1, 1, n);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", chx + 1);
    if (chx[1] == 'A') {
      scanf("%d %d", &x1, &x2);
      printf("%d\n", sgt.query(x1, x2 - 1));
    } else {
      scanf("%d %d", &x1, &x2);
      tim[x1] = x2;
      sgt.edit(1, x1, x2);
    }
  }
  return 0;
}
