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 N = 1e5;
const int MX = 60;
const int BLOCK = 60;
int ar[N + 1];
int dp[N + 1][MX + 1];
int n;
bool pending[N / BLOCK + 1];
inline void memo(int l, int r) {
  for (int i = r; i >= l; --i) {
    for (int j = 0; j < MX; ++j) {
      if (i == r) {
        dp[i][j] = (j == (j / ar[i] * ar[i])) + 1;
      } else {
        int nx = j + 1;
        int tm = 1;
        if (j / ar[i] * ar[i] == j) nx++, tm++;
        if (nx >= 60) nx -= 60;
        dp[i][j] = dp[i + 1][nx] + tm;
      }
    }
  }
}
inline void pre() {
  for (int i = 0; i < n; i += BLOCK) {
    int nx = min(i + BLOCK - 1, n - 1);
    memo(i, nx);
  }
}
int getint() {
  char x = getchar();
  while (x > '9' || x < '0') x = getchar();
  int res = 0;
  for (; '0' <= x && x <= '9'; x = getchar()) {
    res *= 10;
    res += x - '0';
  }
  return res;
}
char getchr() {
  char x = getchar();
  while (x != 'C' && x != 'A') x = getchar();
  return x;
}
int main() {
  n = getint();
  for (int i = 0; i < n; ++i) ar[i] = getint();
  pre();
  int q;
  q = getint();
  while (q--) {
    char x;
    x = getchr();
    int a = getint();
    int b = getint();
    if (x == 'C') {
      a--;
      int startBlock = a / BLOCK * BLOCK;
      pending[a / BLOCK] = true;
      ar[a] = b;
    } else {
      a--, b--;
      if (a / BLOCK * BLOCK == b / BLOCK * BLOCK) {
        int ans = 0;
        while (a < b) {
          if (ans == ans / ar[a] * ar[a])
            ans += 2;
          else
            ans++;
          a++;
        }
        cout << ans << endl;
      } else {
        int aBlock = a / BLOCK * BLOCK;
        if (pending[a / BLOCK]) {
          memo(aBlock, min(aBlock + BLOCK - 1, n - 1));
          pending[a / BLOCK] = false;
        }
        int ans = dp[a][0];
        int cur = dp[a][0] % MX;
        aBlock += BLOCK;
        int nxBlock = min(aBlock + BLOCK - 1, n);
        while (nxBlock < b) {
          if (pending[aBlock / BLOCK]) {
            memo(aBlock, min(aBlock + BLOCK - 1, n - 1));
            pending[aBlock / BLOCK] = false;
          }
          ans += dp[aBlock][cur];
          cur += dp[aBlock][cur];
          cur %= MX;
          aBlock += BLOCK;
          nxBlock = min(aBlock + BLOCK - 1, n);
        }
        while (aBlock < b) {
          if (ans == ans / ar[aBlock] * ar[aBlock])
            ans += 2;
          else
            ans++;
          aBlock++;
        }
        cout << ans << endl;
      }
    }
  }
}
