#include <bits/stdc++.h>
using namespace std;
int a[1000000 + 5];
char b[1000000 + 5];
int las[1000000 + 5];
void get_prime() {
  int n = 1000000;
  for (int i = 2; i <= n; i++)
    if (!las[i]) {
      las[i] = i;
      if (i > 1000) continue;
      for (int j = i * i; j <= n; j += i) las[j] = i;
    }
}
int pre[1000000 + 5];
vector<int> p[1000000 + 5];
void add(int x, int id) {
  pre[id] = id;
  while (x != 1) p[las[x]].push_back(id), x /= las[x];
}
void del(int x, int id) {
  pre[id] = id;
  while (x != 1) {
    if (!p[las[x]].size()) {
      pre[id] = 0;
      return;
    }
    pre[id] = min(pre[id], p[las[x]].back());
    p[las[x]].pop_back();
    x /= las[x];
  }
}
vector<pair<int, int> > Q;
int main() {
  int n, i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%s", b + 1);
  get_prime();
  for (i = 1; i <= n; i++)
    if (b[i] == '*')
      add(a[i], i);
    else
      del(a[i], i);
  long long ans = 0;
  for (i = n; i >= 1; i--) {
    int cnt = 1;
    while (!Q.empty() && Q.back().first >= pre[i])
      cnt += Q.back().second, Q.pop_back();
    Q.push_back(make_pair(pre[i], cnt));
    if (pre[i] == i) ans += cnt;
  }
  printf("%lld\n", ans);
}

### 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;
int a[1000000 + 5];
char b[1000000 + 5];
int las[1000000 + 5];
void get_prime() {
  int n = 1000000;
  for (int i = 2; i <= n; i++)
    if (!las[i]) {
      las[i] = i;
      if (i > 1000) continue;
      for (int j = i * i; j <= n; j += i) las[j] = i;
    }
}
int pre[1000000 + 5];
vector<int> p[1000000 + 5];
void add(int x, int id) {
  pre[id] = id;
  while (x != 1) p[las[x]].push_back(id), x /= las[x];
}
void del(int x, int id) {
  pre[id] = id;
  while (x != 1) {
    if (!p[las[x]].size()) {
      pre[id] = 0;
      return;
    }
    pre[id] = min(pre[id], p[las[x]].back());
    p[las[x]].pop_back();
    x /= las[x];
  }
}
vector<pair<int, int> > Q;
int main() {
  int n, i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%s", b + 1);
  get_prime();
  for (i = 1; i <= n; i++)
    if (b[i] == '*')
      add(a[i], i);
    else
      del(a[i], i);
  long long ans = 0;
  for (i = n; i >= 1; i--) {
    int cnt = 1;
    while (!Q.empty() && Q.back().first >= pre[i])
      cnt += Q.back().second, Q.pop_back();
    Q.push_back(make_pair(pre[i], cnt));
    if (pre[i] == i) ans += cnt;
  }
  printf("%lld\n", ans);
}

```