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>
int v[100007];
int tot;
struct Tree {
  int a[60];
  int l, r, ls, rs;
} tree[100007 << 1];
int pos[100007], father[100007 << 1];
void push(int t) {
  int i;
  if (tree[t].ls) {
    if (tree[t].rs) {
      int mid = (tree[t].l + tree[t].r) >> 1;
      for (i = 0; i < 60; ++i) {
        int t1 = i + tree[tree[t].ls].a[i];
        int t2 = t1 + 1 + (t1 % v[mid] == 0 ? 1 : 0);
        t2 = tree[tree[t].rs].a[t2 % 60];
        tree[t].a[i] = t1 - i + t2 + 1 + (t1 % v[mid] == 0 ? 1 : 0);
      }
    } else {
      for (i = 0; i < 60; ++i) tree[t].a[i] = tree[tree[t].ls].a[i];
    }
  } else if (tree[t].rs) {
    for (i = 0; i < 60; ++i) tree[t].a[i] = tree[tree[t].rs].a[i];
  }
}
void buildtree(int l, int r) {
  int cnt = ++tot;
  tree[cnt].l = l, tree[cnt].r = r;
  tree[cnt].ls = tree[cnt].rs = 0;
  if (l == r) {
    pos[l] = cnt;
    for (int i = 0; i < 60; ++i) tree[cnt].a[i] = 0;
    return;
  }
  int mid = (l + r) >> 1;
  if (l <= mid) {
    father[1 + tot] = cnt;
    tree[cnt].ls = 1 + tot;
    buildtree(l, mid);
  }
  ++mid;
  if (mid <= r) {
    father[1 + tot] = cnt;
    tree[cnt].rs = 1 + tot;
    buildtree(mid, r);
  }
  push(cnt);
}
int query(int t, int l, int r, int time) {
  if (t <= 0) return 0;
  if (l == tree[t].l && r == tree[t].r) return tree[t].a[time % 60];
  int mid = (tree[t].l + tree[t].r) >> 1;
  if (r <= mid) return query(tree[t].ls, l, r, time);
  if (l > mid) return query(tree[t].rs, l, r, time);
  int t1 = time + query(tree[t].ls, l, mid, time);
  int t2 = t1 + 1 + (t1 % v[mid] == 0 ? 1 : 0);
  t2 = query(tree[t].rs, mid + 1, r, t2);
  return t1 - time + 1 + (t1 % v[mid] == 0 ? 1 : 0) + t2;
}
void update(int x, int y) {
  v[x] = y, x = pos[x];
  while (father[x]) {
    x = father[x];
    push(x);
  }
}
int main() {
  char ch;
  int n, q;
  int i, x, y;
  int ANS;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) scanf("%d", &v[i]);
  tot = 0;
  buildtree(1, n);
  scanf("%d", &q);
  while (q--) {
    scanf(" %c %d %d", &ch, &x, &y);
    if (ch == 'A') {
      ANS = query(1, x, y, 0);
      printf("%d\n", ANS);
    } else {
      update(x, y);
    }
  }
  return 0;
}
