#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool f = false;
  for (; !isdigit(c); c = getchar()) f |= c == '-';
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  if (f) x = -x;
}
template <typename T, typename... Arg>
inline void read(T &x, Arg &...args) {
  read(x);
  read(args...);
}
template <typename T>
inline void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 1e6 + 100;
int n, a[N];
char c[N];
int pr[N];
vector<int> v[N];
stack<int> s;
signed main() {
  for (int i = 2; i < N; i++)
    if (!pr[i])
      for (int j = 1; i * j < N; j++) pr[i * j] = i;
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  fgets(c + 1, 998244353, stdin);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    s.push(i);
    while (a[i] > 1) {
      int j = pr[a[i]];
      a[i] /= j;
      if (c[i] == '*')
        v[j].push_back(i);
      else {
        while (s.size() && (!v[j].size() || s.top() > v[j].back())) s.pop();
        if (v[j].size()) v[j].pop_back();
      }
    }
    ans += s.size();
  }
  write(ans);
}

### Prompt

Please create a solution in CPP to 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>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool f = false;
  for (; !isdigit(c); c = getchar()) f |= c == '-';
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  if (f) x = -x;
}
template <typename T, typename... Arg>
inline void read(T &x, Arg &...args) {
  read(x);
  read(args...);
}
template <typename T>
inline void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 1e6 + 100;
int n, a[N];
char c[N];
int pr[N];
vector<int> v[N];
stack<int> s;
signed main() {
  for (int i = 2; i < N; i++)
    if (!pr[i])
      for (int j = 1; i * j < N; j++) pr[i * j] = i;
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  fgets(c + 1, 998244353, stdin);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    s.push(i);
    while (a[i] > 1) {
      int j = pr[a[i]];
      a[i] /= j;
      if (c[i] == '*')
        v[j].push_back(i);
      else {
        while (s.size() && (!v[j].size() || s.top() > v[j].back())) s.pop();
        if (v[j].size()) v[j].pop_back();
      }
    }
    ans += s.size();
  }
  write(ans);
}

```