#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> fac[1000005];
bool comp[1000005];
int n;
int arr[1000005];
vector<pair<int, int>> pos[1000005];
string s;
map<int, int> good[1000005];
long long ans = 0;
vector<int> p;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  for (int i = 2; i <= 1000000; i++) {
    if (comp[i]) {
      continue;
    }
    for (int j = i; j <= 1000000; j += i) {
      comp[j] = true;
      int x = j, cnt = 0;
      while (x % i == 0) {
        x /= i;
        cnt++;
      }
      fac[j].push_back({i, cnt});
    }
  }
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    for (pair<int, int> j : fac[arr[i]]) {
      pos[j.first].push_back({i, j.second});
    }
  }
  cin >> s;
  for (int i = 2; i <= 1000000; i++) {
    vector<int> k = {0};
    for (int j = 0; j < pos[i].size(); j++) {
      int x = pos[i][j].first;
      if (s[x - 1] == '*') {
        for (int it = 1; it <= pos[i][j].second; it++) {
          k.push_back(x);
        }
      } else {
        for (int it = 2; it <= pos[i][j].second and k.size() > 1; it++) {
          k.pop_back();
        }
        good[i][x] = k[k.size() - 1];
        if (k.size() > 1) {
          k.pop_back();
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    p.push_back(i);
    if (s[i - 1] == '/') {
      int x = 2e9;
      for (pair<int, int> j : fac[arr[i]]) {
        x = min(x, good[j.first][i]);
      }
      while (p.size() > 0 and p[p.size() - 1] > x) {
        p.pop_back();
      }
    }
    long long siz = p.size();
    ans += siz;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Generate 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;
vector<pair<int, int>> fac[1000005];
bool comp[1000005];
int n;
int arr[1000005];
vector<pair<int, int>> pos[1000005];
string s;
map<int, int> good[1000005];
long long ans = 0;
vector<int> p;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  for (int i = 2; i <= 1000000; i++) {
    if (comp[i]) {
      continue;
    }
    for (int j = i; j <= 1000000; j += i) {
      comp[j] = true;
      int x = j, cnt = 0;
      while (x % i == 0) {
        x /= i;
        cnt++;
      }
      fac[j].push_back({i, cnt});
    }
  }
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    for (pair<int, int> j : fac[arr[i]]) {
      pos[j.first].push_back({i, j.second});
    }
  }
  cin >> s;
  for (int i = 2; i <= 1000000; i++) {
    vector<int> k = {0};
    for (int j = 0; j < pos[i].size(); j++) {
      int x = pos[i][j].first;
      if (s[x - 1] == '*') {
        for (int it = 1; it <= pos[i][j].second; it++) {
          k.push_back(x);
        }
      } else {
        for (int it = 2; it <= pos[i][j].second and k.size() > 1; it++) {
          k.pop_back();
        }
        good[i][x] = k[k.size() - 1];
        if (k.size() > 1) {
          k.pop_back();
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    p.push_back(i);
    if (s[i - 1] == '/') {
      int x = 2e9;
      for (pair<int, int> j : fac[arr[i]]) {
        x = min(x, good[j.first][i]);
      }
      while (p.size() > 0 and p[p.size() - 1] > x) {
        p.pop_back();
      }
    }
    long long siz = p.size();
    ans += siz;
  }
  cout << ans << "\n";
  return 0;
}

```