#include <bits/stdc++.h>
using namespace std;
template <typename T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using MaxHeap = priority_queue<T>;
const long double PI = 3.14159265358979323846;
const long double E = 2.718281828459;
long long MOD = 1e9 + 7;
long long M(long long n) { return (n % MOD + MOD) % MOD; }
long long pow2(long long n) { return 1ll << n; }
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
const long long MX = 1e6 + 2;
const long long MAXN = MX;
vector<int> prime;
bool is_composite[MAXN];
vector<int> minDiv(MAXN);
void sieve(long long n) {
  std::fill(is_composite, is_composite + n, false);
  for (long long i = 2; i < n; ++i) {
    if (!is_composite[i]) {
      prime.push_back(i);
      minDiv[i] = i;
    }
    for (long long j = 0; j < prime.size() && i * prime[j] < n; ++j) {
      is_composite[i * prime[j]] = true;
      minDiv[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
unordered_map<int, stack<int>> suf;
int main(int argn, char **argv) {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  sieve(MX);
  ;
  int T = 1;
  for (long long(I) = (0); (I) <= ((T)-1); (I)++) {
    long long n;
    cin >> n;
    vector<int> a(n);
    for (long long(i) = (0); (i) <= ((n)-1); (i)++) cin >> a[i];
    string s;
    cin >> s;
    long long ans = 0;
    set<int> lim;
    lim.emplace(n);
    for (long long(i) = (n - 1); (i) >= (0); (i)--) {
      if (s[i] == '/' && a[i] > 1) lim.emplace(i);
      long long tmp = a[i];
      while (tmp > 1) {
        long long pr = minDiv[tmp], c = 0;
        while (minDiv[tmp] == pr) {
          tmp /= pr, c++;
        }
        if (s[i] == '*') {
          while (!suf[pr].empty() && c) {
            auto idx = suf[pr].top();
            c--, a[idx] /= pr;
            if (a[idx] % pr) suf[pr].pop();
            if (a[idx] == 1) lim.erase(idx);
          }
        } else {
          suf[pr].emplace(i);
        }
      };
      long long maxPos = *lim.begin(), earn = maxPos - i;
      ;
      ans += earn;
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
template <typename T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using MaxHeap = priority_queue<T>;
const long double PI = 3.14159265358979323846;
const long double E = 2.718281828459;
long long MOD = 1e9 + 7;
long long M(long long n) { return (n % MOD + MOD) % MOD; }
long long pow2(long long n) { return 1ll << n; }
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
const long long MX = 1e6 + 2;
const long long MAXN = MX;
vector<int> prime;
bool is_composite[MAXN];
vector<int> minDiv(MAXN);
void sieve(long long n) {
  std::fill(is_composite, is_composite + n, false);
  for (long long i = 2; i < n; ++i) {
    if (!is_composite[i]) {
      prime.push_back(i);
      minDiv[i] = i;
    }
    for (long long j = 0; j < prime.size() && i * prime[j] < n; ++j) {
      is_composite[i * prime[j]] = true;
      minDiv[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
unordered_map<int, stack<int>> suf;
int main(int argn, char **argv) {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  sieve(MX);
  ;
  int T = 1;
  for (long long(I) = (0); (I) <= ((T)-1); (I)++) {
    long long n;
    cin >> n;
    vector<int> a(n);
    for (long long(i) = (0); (i) <= ((n)-1); (i)++) cin >> a[i];
    string s;
    cin >> s;
    long long ans = 0;
    set<int> lim;
    lim.emplace(n);
    for (long long(i) = (n - 1); (i) >= (0); (i)--) {
      if (s[i] == '/' && a[i] > 1) lim.emplace(i);
      long long tmp = a[i];
      while (tmp > 1) {
        long long pr = minDiv[tmp], c = 0;
        while (minDiv[tmp] == pr) {
          tmp /= pr, c++;
        }
        if (s[i] == '*') {
          while (!suf[pr].empty() && c) {
            auto idx = suf[pr].top();
            c--, a[idx] /= pr;
            if (a[idx] % pr) suf[pr].pop();
            if (a[idx] == 1) lim.erase(idx);
          }
        } else {
          suf[pr].emplace(i);
        }
      };
      long long maxPos = *lim.begin(), earn = maxPos - i;
      ;
      ans += earn;
    }
    cout << ans << endl;
  }
  return 0;
}

```