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 ST = 18;
const int LEVEL = (1 << ST);
const int SIZE = (LEVEL << 1);
struct Tree {
  int val[SIZE][60];
  void upd(int v) {
    int ch1 = 2 * v;
    int ch2 = 2 * v + 1;
    for (int i = 0; i < 60; i++) {
      int ni = (i + val[ch1][i]) % 60;
      val[v][i] = val[ch1][i] + val[ch2][ni];
    }
  }
  void init(int a[], int n) {
    for (int i = 0; i < 60; i++) {
      for (int v = 0; v < n; v++) {
        if (i % a[v] == 0)
          val[LEVEL + v][i] = 2;
        else
          val[LEVEL + v][i] = 1;
      }
    }
    for (int i = LEVEL - 1; i >= 1; i--) {
      upd(i);
    }
  }
  int getAns(int v, int l, int r, int tl, int tr, int ct) {
    if (tr <= l || tl >= r) return 0;
    if (l <= tl && tr <= r) {
      return val[v][ct % 60];
    }
    int tm = (tl + tr) / 2;
    int ans1 = getAns(2 * v, l, r, tl, tm, ct);
    int ans2 = getAns(2 * v + 1, l, r, tm, tr, ct + ans1);
    return ans1 + ans2;
  }
  void setVal(int pos, int x) {
    for (int i = 0; i < 60; i++) {
      if (i % x == 0)
        val[LEVEL + pos][i] = 2;
      else
        val[LEVEL + pos][i] = 1;
    }
    int v = pos + LEVEL;
    v /= 2;
    while (v) {
      upd(v);
      v /= 2;
    }
  }
};
int a[LEVEL];
Tree tf;
int getAns(int x, int y) { return tf.getAns(1, x, y, 0, LEVEL, 0); }
void setVal(int x, int val) { tf.setVal(x, val); }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  tf.init(a, n);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int x, y;
    char s[2];
    scanf("%s%d%d", s, &x, &y);
    if (s[0] == 'A') {
      x--;
      y--;
      int ans = getAns(x, y);
      printf("%d\n", ans);
    } else {
      x--;
      setVal(x, y);
    }
  }
  return 0;
}
