#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
long long n, a[N], s[N], ans;
vector<long long> v[N], p[N];
char str[N];
void pre() {
  for (long long i = 2; i < N; i++) {
    if (v[i].empty()) {
      for (long long j = i; j < N; j *= i) {
        for (long long k = j; k < N; k += j) {
          v[k].push_back(i);
        }
      }
    }
  }
}
bool ok(int j) {
  if (s[0]) {
    if (p[j].empty()) return true;
    if (p[j].back() < s[s[0]]) return true;
  }
  return false;
}
int main() {
  pre();
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  scanf("%s", str + 1);
  for (int i = 1; i <= n; i++) {
    s[++s[0]] = i;
    for (auto &j : v[a[i]]) {
      if (str[i] == '*') {
        p[j].push_back(i);
      } else {
        while (ok(j)) {
          s[0]--;
        }
        if (!p[j].empty()) p[j].pop_back();
      }
    }
    ans += s[0];
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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 N = 1e6 + 10;
long long n, a[N], s[N], ans;
vector<long long> v[N], p[N];
char str[N];
void pre() {
  for (long long i = 2; i < N; i++) {
    if (v[i].empty()) {
      for (long long j = i; j < N; j *= i) {
        for (long long k = j; k < N; k += j) {
          v[k].push_back(i);
        }
      }
    }
  }
}
bool ok(int j) {
  if (s[0]) {
    if (p[j].empty()) return true;
    if (p[j].back() < s[s[0]]) return true;
  }
  return false;
}
int main() {
  pre();
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  scanf("%s", str + 1);
  for (int i = 1; i <= n; i++) {
    s[++s[0]] = i;
    for (auto &j : v[a[i]]) {
      if (str[i] == '*') {
        p[j].push_back(i);
      } else {
        while (ok(j)) {
          s[0]--;
        }
        if (!p[j].empty()) p[j].pop_back();
      }
    }
    ans += s[0];
  }
  printf("%lld\n", ans);
  return 0;
}

```