#include <bits/stdc++.h>
using namespace std;
const int m = 1000001;
int n;
int a[m];
vector<int> b[m], v[m];
stack<int> second;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 2; i < m; i++) {
    if (b[i].empty()) {
      for (long long j = i; j < m; j *= i) {
        for (long long l = j; l < m; l += j) {
          b[l].push_back(i);
        }
      }
    }
  }
  long long ret = 0;
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    second.push(i);
    for (int j : b[a[i]]) {
      if (c == '*') {
        v[j].push_back(i);
      } else {
        while (!second.empty() &&
               (v[j].empty() ? 1 : second.top() > v[j].back())) {
          second.pop();
        }
        if (!v[j].empty()) v[j].pop_back();
      }
    }
    ret += second.size();
  }
  cout << ret << '\n';
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int m = 1000001;
int n;
int a[m];
vector<int> b[m], v[m];
stack<int> second;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 2; i < m; i++) {
    if (b[i].empty()) {
      for (long long j = i; j < m; j *= i) {
        for (long long l = j; l < m; l += j) {
          b[l].push_back(i);
        }
      }
    }
  }
  long long ret = 0;
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    second.push(i);
    for (int j : b[a[i]]) {
      if (c == '*') {
        v[j].push_back(i);
      } else {
        while (!second.empty() &&
               (v[j].empty() ? 1 : second.top() > v[j].back())) {
          second.pop();
        }
        if (!v[j].empty()) v[j].pop_back();
      }
    }
    ret += second.size();
  }
  cout << ret << '\n';
  return 0;
}

```