Kuzya started going to school. He was given math homework in which he was given an array a of length n and an array of symbols b of length n, consisting of symbols '*' and '/'.

Let's denote a path of calculations for a segment [l; r] (1 ≤ l ≤ r ≤ n) in the following way: 

  * Let x=1 initially. For every i from l to r we will consequently do the following: if b_i= '*', x=x*a_i, and if b_i= '/', then x=(x)/(a_i). Let's call a path of calculations for the segment [l; r] a list of all x that we got during the calculations (the number of them is exactly r - l + 1). 



For example, let a=[7, 12, 3, 5, 4, 10, 9], b=[/, *, /, /, /, *, *], l=2, r=6, then the path of calculations for that segment is [12, 4, 0.8, 0.2, 2].

Let's call a segment [l;r] simple if the path of calculations for it contains only integer numbers. 

Kuzya needs to find the number of simple segments [l;r] (1 ≤ l ≤ r ≤ n). Since he obviously has no time and no interest to do the calculations for each option, he asked you to write a program to get to find that number!

Input

The first line contains a single integer n (2 ≤ n ≤ 10^6).

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).

The third line contains n symbols without spaces between them — the array b_1, b_2 … b_n (b_i= '/' or b_i= '*' for every 1 ≤ i ≤ n).

Output

Print a single integer — the number of simple segments [l;r].

Examples

Input


3
1 2 3
*/*


Output


2


Input


7
6 4 10 1 2 15 1
*/*/*//


Output


8
#include <bits/stdc++.h>
using namespace std;
template <typename S, typename T>
bool minx(S &a, const T &b) {
  return a < b ? 1 : a = b, 0;
}
template <typename T>
void int_in(T &res) {
  int flag = 1;
  char ch;
  if ((ch = getchar()) == '-')
    flag = -1;
  else if (ch >= '0' && ch <= '9')
    res = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + ch - '0';
  res *= flag;
}
template <typename T>
void int_out(T a) {
  if (a < 0) {
    putchar('-');
    a = -a;
  }
  if (a > 9) int_out(a / 10);
  putchar(a % 10 + '0');
}
const int INF = 0x3f3f3f3f;
const double tol = 1e-8;
const int N = 100005;
int a[N];
int main() {
  int t;
  int_in(t);
  while (t--) {
    int n;
    int_in(n);
    for (int i = 1; i <= n; ++i) {
      int_in(a[i]);
    }
    if (n & 1) {
      if (a[1] + a[2] != 0) {
        printf("%d %d %d ", a[3], a[3], -(a[1] + a[2]));
      } else if (a[1] + a[3] != 0) {
        printf("%d %d %d ", a[2], -(a[1] + a[3]), a[2]);
      } else {
        printf("%d %d %d ", -(a[2] + a[3]), a[1], a[1]);
      }
    }
    for (int i = 1 + 3 * (n & 1); i <= n; i += 2) {
      int_out(-a[i + 1]);
      putchar(' ');
      int_out(a[i]);
      putchar(' ');
    }
    puts("");
  }
  return 0;
}
