#include <bits/stdc++.h>
using namespace std;
const long long N = 1000009;
const long long INF = 0x3f3f3f3f;
long long dta[N];
long long L[N];
char sing[N];
long long prime[N];
vector<long long> pos[N];
signed main(void) {
  for (long long i = 2; i < N; i++)
    if (prime[i] == 0) {
      prime[i] = i;
      for (long long j = i * i; j < N; j += i) prime[j] = i;
    }
  long long n;
  long long temp;
  cin >> n;
  for (long long i = 1; i <= n; i++) L[i] = INF;
  for (long long i = 1; i <= n; i++) cin >> dta[i];
  cin >> sing + 1;
  for (long long i = 1; i <= n; i++) {
    L[i] = i;
    if (sing[i] == '*') {
      temp = dta[i];
      while (temp > 1) {
        pos[prime[temp]].push_back(i);
        temp /= prime[temp];
      }
    } else {
      temp = dta[i];
      while (temp > 1) {
        if (pos[prime[temp]].empty()) {
          L[i] = -1;
          break;
        }
        L[i] = min(L[i], pos[prime[temp]].back());
        pos[prime[temp]].pop_back();
        temp /= prime[temp];
      }
    }
  }
  long long ans = 0;
  vector<long long> vec;
  for (long long i = 1; i <= n; i++) {
    while (!vec.empty() && vec.back() > L[i]) vec.pop_back();
    if (i == L[i]) vec.push_back(i);
    ans += vec.size();
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 long long N = 1000009;
const long long INF = 0x3f3f3f3f;
long long dta[N];
long long L[N];
char sing[N];
long long prime[N];
vector<long long> pos[N];
signed main(void) {
  for (long long i = 2; i < N; i++)
    if (prime[i] == 0) {
      prime[i] = i;
      for (long long j = i * i; j < N; j += i) prime[j] = i;
    }
  long long n;
  long long temp;
  cin >> n;
  for (long long i = 1; i <= n; i++) L[i] = INF;
  for (long long i = 1; i <= n; i++) cin >> dta[i];
  cin >> sing + 1;
  for (long long i = 1; i <= n; i++) {
    L[i] = i;
    if (sing[i] == '*') {
      temp = dta[i];
      while (temp > 1) {
        pos[prime[temp]].push_back(i);
        temp /= prime[temp];
      }
    } else {
      temp = dta[i];
      while (temp > 1) {
        if (pos[prime[temp]].empty()) {
          L[i] = -1;
          break;
        }
        L[i] = min(L[i], pos[prime[temp]].back());
        pos[prime[temp]].pop_back();
        temp /= prime[temp];
      }
    }
  }
  long long ans = 0;
  vector<long long> vec;
  for (long long i = 1; i <= n; i++) {
    while (!vec.empty() && vec.back() > L[i]) vec.pop_back();
    if (i == L[i]) vec.push_back(i);
    ans += vec.size();
  }
  cout << ans << endl;
  return 0;
}

```