#include <bits/stdc++.h>
using namespace std;
int const maxn = 1e6 + 5;
int a[maxn];
int prime[maxn];
int L[maxn];
vector<int> pos[maxn];
inline void add(int x, int l) {
  L[l] = l;
  while (x > 1) {
    pos[prime[x]].push_back(l);
    x /= prime[x];
  }
}
inline void del(int x, int l) {
  if (x == 1) {
    L[l] = l;
    return;
  }
  L[l] = l;
  while (x > 1) {
    if ((int)pos[prime[x]].size() == 0) {
      L[l] = 0;
      return;
    }
    L[l] = min(L[l], pos[prime[x]].back());
    pos[prime[x]].pop_back();
    x /= prime[x];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 2; i < maxn; ++i) {
    if (prime[i] == 0) {
      prime[i] = i;
      if (i > 1000) continue;
      for (int j = i * i; j < maxn; j += i) {
        prime[j] = i;
      }
    }
  }
  char type;
  for (int i = 1; i <= n; ++i) {
    cin >> type;
    if (type == '*')
      add(a[i], i);
    else
      del(a[i], i);
  }
  long long answer = 0;
  vector<pair<int, int> > f_min;
  for (int i = n; i >= 1; --i) {
    int cnt = 1;
    while ((int)f_min.size() > 0 && f_min.back().first >= i) {
      cnt += f_min.back().second;
      f_min.pop_back();
    }
    f_min.push_back({L[i], cnt});
    if (L[i] == i) answer += (long long)cnt;
  }
  cout << answer << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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;
int const maxn = 1e6 + 5;
int a[maxn];
int prime[maxn];
int L[maxn];
vector<int> pos[maxn];
inline void add(int x, int l) {
  L[l] = l;
  while (x > 1) {
    pos[prime[x]].push_back(l);
    x /= prime[x];
  }
}
inline void del(int x, int l) {
  if (x == 1) {
    L[l] = l;
    return;
  }
  L[l] = l;
  while (x > 1) {
    if ((int)pos[prime[x]].size() == 0) {
      L[l] = 0;
      return;
    }
    L[l] = min(L[l], pos[prime[x]].back());
    pos[prime[x]].pop_back();
    x /= prime[x];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 2; i < maxn; ++i) {
    if (prime[i] == 0) {
      prime[i] = i;
      if (i > 1000) continue;
      for (int j = i * i; j < maxn; j += i) {
        prime[j] = i;
      }
    }
  }
  char type;
  for (int i = 1; i <= n; ++i) {
    cin >> type;
    if (type == '*')
      add(a[i], i);
    else
      del(a[i], i);
  }
  long long answer = 0;
  vector<pair<int, int> > f_min;
  for (int i = n; i >= 1; --i) {
    int cnt = 1;
    while ((int)f_min.size() > 0 && f_min.back().first >= i) {
      cnt += f_min.back().second;
      f_min.pop_back();
    }
    f_min.push_back({L[i], cnt});
    if (L[i] == i) answer += (long long)cnt;
  }
  cout << answer << '\n';
  return 0;
}

```