#include <bits/stdc++.h>
using namespace std;
const int ma = 1e6, inf = 1e9;
int n, nn[1000069], fq[1000069], a[1000069];
pair<int, int> fz[1000069][7];
vector<pair<int, int>> sk[1000069];
bitset<1000069> spc;
multiset<int> ms;
int main() {
  int i, j, k, w, c, u, ww;
  long long z = 0;
  char ch;
  for (i = 2; i <= ma; i++) {
    if (!fq[i]) {
      for (j = i; j <= ma; j += i) {
        for (c = 0, k = j; k % i == 0; k /= i, c++)
          ;
        fz[j][fq[j]] = {i, c};
        fq[j]++;
      }
      sk[i].push_back({-1, -inf});
      sk[i].push_back({0, 0});
      nn[i] = 1;
      ms.insert(-1);
    }
  }
  scanf("%d", &n);
  for (i = n; i; i--) {
    scanf("%d", a + i);
  }
  for (i = n; i; i--) {
    scanf(" %c", &ch);
    spc[i] = ch == '/';
  }
  for (i = 1; i <= n; i++) {
    u = spc[i] * 2 - 1;
    for (j = 0; j < fq[a[i]]; j++) {
      k = fz[a[i]][j].first;
      w = fz[a[i]][j].second;
      sk[k][nn[k]].first = i - 1;
      ms.erase(ms.find(sk[k][nn[k] - 1].first));
      ww = sk[k][nn[k]].second + w * u;
      for (; nn[k] && sk[k][nn[k]].second >= ww; sk[k].pop_back(), nn[k]--)
        ;
      ms.insert(sk[k][nn[k]].first);
      sk[k].push_back({i, ww});
      nn[k]++;
    }
    z += i - 1 - *prev(ms.end());
  }
  printf("%lld\n", z);
}

### Prompt

Please provide a CPP coded solution 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 int ma = 1e6, inf = 1e9;
int n, nn[1000069], fq[1000069], a[1000069];
pair<int, int> fz[1000069][7];
vector<pair<int, int>> sk[1000069];
bitset<1000069> spc;
multiset<int> ms;
int main() {
  int i, j, k, w, c, u, ww;
  long long z = 0;
  char ch;
  for (i = 2; i <= ma; i++) {
    if (!fq[i]) {
      for (j = i; j <= ma; j += i) {
        for (c = 0, k = j; k % i == 0; k /= i, c++)
          ;
        fz[j][fq[j]] = {i, c};
        fq[j]++;
      }
      sk[i].push_back({-1, -inf});
      sk[i].push_back({0, 0});
      nn[i] = 1;
      ms.insert(-1);
    }
  }
  scanf("%d", &n);
  for (i = n; i; i--) {
    scanf("%d", a + i);
  }
  for (i = n; i; i--) {
    scanf(" %c", &ch);
    spc[i] = ch == '/';
  }
  for (i = 1; i <= n; i++) {
    u = spc[i] * 2 - 1;
    for (j = 0; j < fq[a[i]]; j++) {
      k = fz[a[i]][j].first;
      w = fz[a[i]][j].second;
      sk[k][nn[k]].first = i - 1;
      ms.erase(ms.find(sk[k][nn[k] - 1].first));
      ww = sk[k][nn[k]].second + w * u;
      for (; nn[k] && sk[k][nn[k]].second >= ww; sk[k].pop_back(), nn[k]--)
        ;
      ms.insert(sk[k][nn[k]].first);
      sk[k].push_back({i, ww});
      nn[k]++;
    }
    z += i - 1 - *prev(ms.end());
  }
  printf("%lld\n", z);
}

```