#include <bits/stdc++.h>
using namespace std;
void solve();
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
const long long N = 1e6 + 100, A = 1e6 + 100, K = 1001, INF = 2e18,
                MOD = 1e9 + 7;
long long n, a[N], g[N], prime[N];
vector<long long> v[N];
string s;
vector<long long> primes(long long e) {
  vector<long long> res;
  while (e > 1) {
    res.push_back(prime[e]);
    e /= prime[e];
  }
  return res;
}
void solve() {
  for (long long i = 2; i < N; i++) {
    if (prime[i] == 0) {
      prime[i] = i;
      for (long long j = i * i; j < N; j += i) {
        prime[j] = i;
      }
    }
  }
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  cin >> s;
  for (long long i = 0; i < n; i++) {
    vector<long long> pr = primes(a[i]);
    if (s[i] == '*') {
      g[i] = i;
      for (long long x : pr) {
        v[x].push_back(i);
      }
    } else {
      long long mn = INF;
      for (long long x : pr) {
        if (v[x].empty()) {
          mn = -1;
        } else {
          mn = min(mn, v[x].back());
          v[x].pop_back();
        }
      }
      g[i] = mn;
    }
  }
  long long ans = 0;
  vector<pair<long long, long long>> u;
  for (long long i = n - 1; i >= 0; i--) {
    u.push_back({g[i], i});
    while (!u.empty() && u.back().first >= i) {
      u.pop_back();
    }
    if (u.empty()) {
      ans += n - i;
    } else {
      ans += u.back().second - i;
    }
  }
  cout << ans;
}

### Prompt

Your task is to create 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;
void solve();
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
const long long N = 1e6 + 100, A = 1e6 + 100, K = 1001, INF = 2e18,
                MOD = 1e9 + 7;
long long n, a[N], g[N], prime[N];
vector<long long> v[N];
string s;
vector<long long> primes(long long e) {
  vector<long long> res;
  while (e > 1) {
    res.push_back(prime[e]);
    e /= prime[e];
  }
  return res;
}
void solve() {
  for (long long i = 2; i < N; i++) {
    if (prime[i] == 0) {
      prime[i] = i;
      for (long long j = i * i; j < N; j += i) {
        prime[j] = i;
      }
    }
  }
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  cin >> s;
  for (long long i = 0; i < n; i++) {
    vector<long long> pr = primes(a[i]);
    if (s[i] == '*') {
      g[i] = i;
      for (long long x : pr) {
        v[x].push_back(i);
      }
    } else {
      long long mn = INF;
      for (long long x : pr) {
        if (v[x].empty()) {
          mn = -1;
        } else {
          mn = min(mn, v[x].back());
          v[x].pop_back();
        }
      }
      g[i] = mn;
    }
  }
  long long ans = 0;
  vector<pair<long long, long long>> u;
  for (long long i = n - 1; i >= 0; i--) {
    u.push_back({g[i], i});
    while (!u.empty() && u.back().first >= i) {
      u.pop_back();
    }
    if (u.empty()) {
      ans += n - i;
    } else {
      ans += u.back().second - i;
    }
  }
  cout << ans;
}

```