#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int a[N];
char s[N];
unordered_map<int, int> d[N];
stack<unordered_map<int, int> *> q;
vector<int> prime;
int vis[1005];
void euler() {
  for (int i = 2; i <= 1000; ++i) {
    if (!vis[i]) prime.push_back(i);
    for (int j = 0; j < prime.size() && i * prime[j] < 1000; ++j) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
void init() {
  for (int i = 1; i <= 1000000; ++i) {
    int x = i;
    for (int j : prime) {
      if (j * j > x) break;
      while (x % j == 0) {
        ++d[i][j];
        x /= j;
      }
    }
    if (x > 1) ++d[i][x];
  }
}
bool check(unordered_map<int, int> &x, unordered_map<int, int> &y) {
  for (auto &i : y) {
    if (x.find(i.first) == x.end() || x[i.first] < i.second) return false;
  }
  return true;
}
void merge(unordered_map<int, int> &x, unordered_map<int, int> &y) {
  for (auto &i : y) {
    x[i.first] += i.second;
  }
}
void sub(unordered_map<int, int> &x, unordered_map<int, int> &y) {
  for (auto &i : y) {
    x[i.first] -= i.second;
    if (!x[i.first]) x.erase(i.first);
  }
}
int main() {
  euler();
  init();
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  scanf("%s", s);
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '*') {
      auto *mp = new unordered_map<int, int>;
      merge(*mp, d[a[i]]);
      q.push(mp);
    } else {
      auto *mp = new unordered_map<int, int>;
      merge(*mp, d[1]);
      q.push(mp);
      unordered_map<int, int> &cur = d[a[i]];
      while (!q.empty() && !check(*q.top(), cur)) {
        unordered_map<int, int> *p = q.top();
        q.pop();
        if (!q.empty()) {
          merge(*p, *q.top());
          delete q.top();
          q.pop();
          q.push(p);
        }
      }
      if (!q.empty()) sub(*q.top(), cur);
    }
    ans += q.size();
  }
  printf("%lld\n", ans);
  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;
const int N = 1e6 + 5;
int a[N];
char s[N];
unordered_map<int, int> d[N];
stack<unordered_map<int, int> *> q;
vector<int> prime;
int vis[1005];
void euler() {
  for (int i = 2; i <= 1000; ++i) {
    if (!vis[i]) prime.push_back(i);
    for (int j = 0; j < prime.size() && i * prime[j] < 1000; ++j) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
void init() {
  for (int i = 1; i <= 1000000; ++i) {
    int x = i;
    for (int j : prime) {
      if (j * j > x) break;
      while (x % j == 0) {
        ++d[i][j];
        x /= j;
      }
    }
    if (x > 1) ++d[i][x];
  }
}
bool check(unordered_map<int, int> &x, unordered_map<int, int> &y) {
  for (auto &i : y) {
    if (x.find(i.first) == x.end() || x[i.first] < i.second) return false;
  }
  return true;
}
void merge(unordered_map<int, int> &x, unordered_map<int, int> &y) {
  for (auto &i : y) {
    x[i.first] += i.second;
  }
}
void sub(unordered_map<int, int> &x, unordered_map<int, int> &y) {
  for (auto &i : y) {
    x[i.first] -= i.second;
    if (!x[i.first]) x.erase(i.first);
  }
}
int main() {
  euler();
  init();
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  scanf("%s", s);
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '*') {
      auto *mp = new unordered_map<int, int>;
      merge(*mp, d[a[i]]);
      q.push(mp);
    } else {
      auto *mp = new unordered_map<int, int>;
      merge(*mp, d[1]);
      q.push(mp);
      unordered_map<int, int> &cur = d[a[i]];
      while (!q.empty() && !check(*q.top(), cur)) {
        unordered_map<int, int> *p = q.top();
        q.pop();
        if (!q.empty()) {
          merge(*p, *q.top());
          delete q.top();
          q.pop();
          q.push(p);
        }
      }
      if (!q.empty()) sub(*q.top(), cur);
    }
    ans += q.size();
  }
  printf("%lld\n", ans);
  return 0;
}

```