#include <bits/stdc++.h>
using namespace std;
int T, n, p[1100000], l[1100000], a[1100000], ans[1100000];
vector<int> v[1100000];
string s;
int main() {
  for (int i = 2; i <= 1000000; i++) {
    if (!p[i]) {
      p[i] = i;
      for (int j = 2; i * j <= 1000000; j++) {
        if (!p[i * j]) p[i * j] = i;
      }
    }
  }
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  cin >> s;
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == '*') {
      l[i] = i;
      while (a[i] != 1) {
        v[p[a[i]]].push_back(i);
        a[i] /= p[a[i]];
      }
    } else {
      l[i] = i;
      while (a[i] != 1) {
        if (v[p[a[i]]].empty())
          l[i] = -1;
        else
          l[i] = min(l[i], v[p[a[i]]].back()), v[p[a[i]]].pop_back();
        a[i] /= p[a[i]];
      }
    }
  }
  stack<int> s;
  long long sum = 0;
  for (int i = n; i >= 1; i--) {
    ans[i] = 1;
    while (!s.empty() && s.top() != -1 && l[s.top()] >= l[i])
      ans[i] += ans[s.top()], s.pop();
    s.push(i);
    if (l[i] == i) sum += ans[i];
  }
  cout << sum << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int T, n, p[1100000], l[1100000], a[1100000], ans[1100000];
vector<int> v[1100000];
string s;
int main() {
  for (int i = 2; i <= 1000000; i++) {
    if (!p[i]) {
      p[i] = i;
      for (int j = 2; i * j <= 1000000; j++) {
        if (!p[i * j]) p[i * j] = i;
      }
    }
  }
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  cin >> s;
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == '*') {
      l[i] = i;
      while (a[i] != 1) {
        v[p[a[i]]].push_back(i);
        a[i] /= p[a[i]];
      }
    } else {
      l[i] = i;
      while (a[i] != 1) {
        if (v[p[a[i]]].empty())
          l[i] = -1;
        else
          l[i] = min(l[i], v[p[a[i]]].back()), v[p[a[i]]].pop_back();
        a[i] /= p[a[i]];
      }
    }
  }
  stack<int> s;
  long long sum = 0;
  for (int i = n; i >= 1; i--) {
    ans[i] = 1;
    while (!s.empty() && s.top() != -1 && l[s.top()] >= l[i])
      ans[i] += ans[s.top()], s.pop();
    s.push(i);
    if (l[i] == i) sum += ans[i];
  }
  cout << sum << endl;
  return 0;
}

```