#include <bits/stdc++.h>
using namespace std;
string _reset = "\u001b[0m", _yellow = "\u001b[33m", _bold = "\u001b[1m";
void DBG() { cerr << "]" << _reset << endl; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << to_string(h);
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
const int INF = 1e9 + 7, N = 1e6 + 7;
int a[N], lp[N];
int n;
multiset<int> ms;
struct DS {
  vector<pair<int, int> > tt;
  int lz, ans;
  DS() { lz = 0, ans = -1; }
  void undo() {
    if (ans != -1) ms.erase(ms.find(ans));
  }
  void godo() {
    if (tt.empty())
      ans = n;
    else
      ans = tt.back().first;
    ms.insert(ans);
  }
  void upd(int pos, int op) {
    undo();
    tt.emplace_back(pos, -lz);
    lz += op;
    while (tt.size() && tt.back().second + lz >= 0) tt.pop_back();
    godo();
  }
} ds[N];
long long ans;
signed main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  for (int i = 2; i < N; i++) {
    if (lp[i]) continue;
    lp[i] = i;
    for (int j = 2 * i; j < N; j += i) lp[j] = i;
  }
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  string s;
  cin >> s;
  ans = 0;
  ms.insert(n);
  auto upd = [&](int x, int pos, int op) {
    while (x != 1) {
      int p = lp[x];
      int p_cnt = 0;
      while (x % p == 0) {
        p_cnt++;
        x /= lp[x];
      }
      ds[p].upd(pos, op * p_cnt);
    }
  };
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] == 1) s[i] = '*';
    if (s[i] == '*')
      upd(a[i], i, 1);
    else
      upd(a[i], i, -1);
    ans += *ms.begin() - i;
  }
  cout << ans << '\n';
}

### 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;
string _reset = "\u001b[0m", _yellow = "\u001b[33m", _bold = "\u001b[1m";
void DBG() { cerr << "]" << _reset << endl; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << to_string(h);
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
const int INF = 1e9 + 7, N = 1e6 + 7;
int a[N], lp[N];
int n;
multiset<int> ms;
struct DS {
  vector<pair<int, int> > tt;
  int lz, ans;
  DS() { lz = 0, ans = -1; }
  void undo() {
    if (ans != -1) ms.erase(ms.find(ans));
  }
  void godo() {
    if (tt.empty())
      ans = n;
    else
      ans = tt.back().first;
    ms.insert(ans);
  }
  void upd(int pos, int op) {
    undo();
    tt.emplace_back(pos, -lz);
    lz += op;
    while (tt.size() && tt.back().second + lz >= 0) tt.pop_back();
    godo();
  }
} ds[N];
long long ans;
signed main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  for (int i = 2; i < N; i++) {
    if (lp[i]) continue;
    lp[i] = i;
    for (int j = 2 * i; j < N; j += i) lp[j] = i;
  }
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  string s;
  cin >> s;
  ans = 0;
  ms.insert(n);
  auto upd = [&](int x, int pos, int op) {
    while (x != 1) {
      int p = lp[x];
      int p_cnt = 0;
      while (x % p == 0) {
        p_cnt++;
        x /= lp[x];
      }
      ds[p].upd(pos, op * p_cnt);
    }
  };
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] == 1) s[i] = '*';
    if (s[i] == '*')
      upd(a[i], i, 1);
    else
      upd(a[i], i, -1);
    ans += *ms.begin() - i;
  }
  cout << ans << '\n';
}

```