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;
vector<int> pr(1000001, 1);
vector<vector<int> > pv(1000001, vector<int>());
int main() {
  for (int i = 2; i < 1000001; i++) {
    if (pr[i] == 1) {
      for (int j = i; j < 1000001; j += i) {
        pr[j] = i;
      }
    }
  }
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> fix(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    fix[i] = i;
  }
  string b;
  cin >> b;
  for (int i = n - 1; i >= 0; i--) {
    if (b[i] == '*') {
      int v = a[i];
      while (v > 1) {
        int p = pr[v];
        v /= p;
        if (!pv[p].empty()) {
          fix[pv[p].back()] = i;
          pv[p].pop_back();
        }
      }
    } else {
      int v = a[i];
      while (v > 1) {
        int p = pr[v];
        v /= p;
        pv[p].push_back(i);
      }
    }
  }
  for (int i = 0; i < 1000001; i++) {
    while (!pv[i].empty()) {
      fix[pv[i].back()] = -1;
      pv[i].pop_back();
    }
  }
  int right = -1;
  long long out = 0;
  for (int left = 0; left < n; left++) {
    while (right < n - 1) {
      if (fix[right + 1] < left) break;
      right++;
    }
    out += right - left + 1;
    if (right < left) right = left;
  }
  cout << out << '\n';
}
