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;
int n;
int a[1 << 17];
int st[1 << 18][60], sv[1 << 18][60];
const int off = 1 << 17;
int bad[60][7] = {};
char c[5];
inline int pMod(int x) {
  if (x >= 60) return x - 60;
  return x;
}
void buildSt() {
  for (int i = n; i < off; ++i) a[i] = 2;
  for (int i = 0; i < (int)(off); ++i)
    for (int j = 0; j < (int)(60); ++j)
      st[off + i][j] = pMod(j + 1 + bad[j][a[i]]),
               sv[off + i][j] = 1 + bad[j][a[i]];
  for (int i = off - 1; i >= 1; --i)
    for (int j = 0; j < (int)(60); ++j) {
      st[i][j] = st[(i << 1) | 1][st[i << 1][j]];
      sv[i][j] = sv[i << 1][j] + sv[(i << 1) | 1][st[i << 1][j]];
    }
}
void updateSt(int pos, int val) {
  if (a[pos] == val) return;
  a[pos] = val;
  for (int j = 0; j < (int)(60); ++j)
    st[off + pos][j] = pMod(j + 1 + bad[j][val]),
             sv[off + pos][j] = 1 + bad[j][val];
  for (int i = (off + pos) >> 1; i >= 1; i >>= 1)
    for (int j = 0; j < (int)(60); ++j) {
      st[i][j] = st[(i << 1) | 1][st[i << 1][j]];
      sv[i][j] = sv[i << 1][j] + sv[(i << 1) | 1][st[i << 1][j]];
    }
}
int state, val;
void querySt(int v, int L, int R, int l, int r) {
  if (l == L && r == R) {
    val += sv[v][state];
    state = st[v][state];
    return;
  }
  int mid = (L + R) >> 1;
  if (l <= mid) querySt(v << 1, L, mid, l, min(r, mid));
  if (r > mid) querySt((v << 1) | 1, mid + 1, R, max(l, mid + 1), r);
}
int main() {
  for (int i = 0; i < (int)(60); ++i)
    for (int j = 2; j <= 6; ++j) bad[i][j] = (i % j == 0);
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); ++i) scanf("%d", a + i);
  buildSt();
  int q;
  scanf("%d", &q);
  for (int i = 0; i < (int)(q); ++i) {
    int x, y;
    scanf("%s%d%d", c, &x, &y);
    if (c[0] == 'A') {
      state = val = 0;
      querySt(1, 0, off - 1, x - 1, y - 2);
      printf("%d\n", val);
    } else {
      updateSt(x - 1, y);
    }
  }
  return 0;
}
