#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int v[N], prime[N], cntp;
inline void init(int n) {
  for (int i = 2; i <= n; i++) {
    if (v[i] == 0) v[i] = i, prime[++cntp] = i;
    for (int j = 1; j <= cntp; j++) {
      if (prime[j] > v[i] || prime[j] > n / i) break;
      v[i * prime[j]] = prime[j];
    }
  }
}
int a[N], n, L[N];
char b[N];
vector<int> pos[N];
vector<pair<int, int>> stk;
inline void add(int i, int x) {
  while (x > 1) pos[v[x]].push_back(i), x /= v[x];
}
inline void del(int i, int x) {
  while (x > 1) {
    if (pos[v[x]].empty()) return L[i] = 0, void();
    L[i] = min(L[i], pos[v[x]].back());
    pos[v[x]].pop_back(), x /= v[x];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  init(*max_element(a + 1, a + 1 + n));
  iota(L + 1, L + 1 + n, 1), cin >> (b + 1);
  for (int i = 1; i <= n; i++) b[i] == '*' ? add(i, a[i]) : del(i, a[i]);
  long long ans = 0;
  for (int i = n; i >= 1; i--) {
    if (!L[i]) {
      stk.clear();
      continue;
    }
    int cnt = 1;
    while (!stk.empty() && stk.back().first >= L[i])
      cnt += stk.back().second, stk.pop_back();
    stk.emplace_back(L[i], cnt);
    if (L[i] == i) ans += cnt;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 + 10;
int v[N], prime[N], cntp;
inline void init(int n) {
  for (int i = 2; i <= n; i++) {
    if (v[i] == 0) v[i] = i, prime[++cntp] = i;
    for (int j = 1; j <= cntp; j++) {
      if (prime[j] > v[i] || prime[j] > n / i) break;
      v[i * prime[j]] = prime[j];
    }
  }
}
int a[N], n, L[N];
char b[N];
vector<int> pos[N];
vector<pair<int, int>> stk;
inline void add(int i, int x) {
  while (x > 1) pos[v[x]].push_back(i), x /= v[x];
}
inline void del(int i, int x) {
  while (x > 1) {
    if (pos[v[x]].empty()) return L[i] = 0, void();
    L[i] = min(L[i], pos[v[x]].back());
    pos[v[x]].pop_back(), x /= v[x];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  init(*max_element(a + 1, a + 1 + n));
  iota(L + 1, L + 1 + n, 1), cin >> (b + 1);
  for (int i = 1; i <= n; i++) b[i] == '*' ? add(i, a[i]) : del(i, a[i]);
  long long ans = 0;
  for (int i = n; i >= 1; i--) {
    if (!L[i]) {
      stk.clear();
      continue;
    }
    int cnt = 1;
    while (!stk.empty() && stk.back().first >= L[i])
      cnt += stk.back().second, stk.pop_back();
    stk.emplace_back(L[i], cnt);
    if (L[i] == i) ans += cnt;
  }
  cout << ans << "\n";
  return 0;
}

```