#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
int t, n, q;
int v[1000010];
string s;
vector<int> pd[1000010];
vector<int> cod[1000010];
int p[1000010], siz;
set<int> ses;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 2; i < 1000010; i++) {
    if (!p[i]) {
      for (int j = i; j < 1000010; j += i) {
        p[j] = 1;
        pd[j].push_back(i);
      }
    }
  }
  cin >> n;
  for (int i = 0; i < n; i++) cin >> v[i];
  cin >> s;
  long long ris = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == '/') {
      for (auto j : pd[v[i]]) {
        cod[j].push_back(i);
        ses.insert(i);
      }
      if (v[i] == 1) {
        int mim = n;
        if (ses.size()) mim = *ses.begin();
        ris += mim - i;
      }
    } else {
      int kek = v[i], mim = n;
      for (auto j : pd[v[i]]) {
        while (kek % j == 0 && cod[j].size()) {
          int k = cod[j].back();
          v[k] /= j;
          if (v[k] == 1) {
            ses.erase(ses.find(k));
          }
          if (v[k] % j) {
            cod[j].pop_back();
          }
          kek /= j;
        }
      }
      if (ses.size()) mim = *ses.begin();
      ris += mim - i;
    }
  }
  cout << ris << "\n";
}

### 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>
#pragma GCC optimize("O3")
using namespace std;
int t, n, q;
int v[1000010];
string s;
vector<int> pd[1000010];
vector<int> cod[1000010];
int p[1000010], siz;
set<int> ses;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 2; i < 1000010; i++) {
    if (!p[i]) {
      for (int j = i; j < 1000010; j += i) {
        p[j] = 1;
        pd[j].push_back(i);
      }
    }
  }
  cin >> n;
  for (int i = 0; i < n; i++) cin >> v[i];
  cin >> s;
  long long ris = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == '/') {
      for (auto j : pd[v[i]]) {
        cod[j].push_back(i);
        ses.insert(i);
      }
      if (v[i] == 1) {
        int mim = n;
        if (ses.size()) mim = *ses.begin();
        ris += mim - i;
      }
    } else {
      int kek = v[i], mim = n;
      for (auto j : pd[v[i]]) {
        while (kek % j == 0 && cod[j].size()) {
          int k = cod[j].back();
          v[k] /= j;
          if (v[k] == 1) {
            ses.erase(ses.find(k));
          }
          if (v[k] % j) {
            cod[j].pop_back();
          }
          kek /= j;
        }
      }
      if (ses.size()) mim = *ses.begin();
      ris += mim - i;
    }
  }
  cout << ris << "\n";
}

```