#include <bits/stdc++.h>
using namespace std;
template <typename S, typename T>
bool minx(S &a, const T &b) {
  return a < b ? 1 : a = b, 0;
}
long long int_in() {
  long long res = 0;
  int ch, flag = 0;
  if ((ch = getchar()) == '-')
    flag = 1;
  else if (ch >= '0' && ch <= '9')
    res = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + ch - '0';
  return flag ? -res : res;
}
void int_out(long long a) {
  if (a > 9) int_out(a / 10);
  putchar(a % 10 + '0');
}
const int INF = 0x3f3f3f3f;
const double tol = 1e-8;
const int N = 1000005;
int primes[N], minp[N], a[N], left_pos[N];
int liner_sieve(int n, int primes[], int minp[]) {
  int cnt = 0;
  memset(minp, 0, n * sizeof *minp);
  for (int i = 2; i < n; ++i) {
    if (!minp[i]) {
      minp[i] = primes[cnt++] = i;
    }
    for (int j = 0; primes[j] * i < n; ++j) {
      minp[primes[j] * i] = primes[j];
      if (i % primes[j] == 0) break;
    }
  }
  return cnt;
}
vector<int> pos[N];
void add(int idx, int num) {
  left_pos[idx] = idx;
  for (int p = minp[num]; num != 1; p = minp[num /= p]) pos[p].push_back(idx);
}
void del(int idx, int num) {
  left_pos[idx] = idx;
  for (int p = minp[num]; num != 1; p = minp[num /= p]) {
    if (pos[p].size() == 0) {
      left_pos[idx] = 0;
      return;
    }
    minx(left_pos[idx], pos[p].back());
    pos[p].pop_back();
  }
}
int main() {
  int n;
  cin >> n;
  liner_sieve(N, primes, minp);
  for (int i = 1; i <= n; ++i) a[i] = int_in();
  char type;
  for (int i = 1; i <= n; ++i) {
    scanf("%c", &type);
    if (type == '*')
      add(i, a[i]);
    else
      del(i, a[i]);
  }
  vector<int> stk;
  left_pos[n + 1] = -1;
  stk.push_back(n + 1);
  long long ans = 0;
  for (int i = n; i >= 1; --i) {
    while (left_pos[stk.back()] >= left_pos[i]) {
      stk.pop_back();
    }
    if (left_pos[i] == i) ans += stk.back() - i;
    stk.push_back(i);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please formulate a cpp solution 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 S, typename T>
bool minx(S &a, const T &b) {
  return a < b ? 1 : a = b, 0;
}
long long int_in() {
  long long res = 0;
  int ch, flag = 0;
  if ((ch = getchar()) == '-')
    flag = 1;
  else if (ch >= '0' && ch <= '9')
    res = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + ch - '0';
  return flag ? -res : res;
}
void int_out(long long a) {
  if (a > 9) int_out(a / 10);
  putchar(a % 10 + '0');
}
const int INF = 0x3f3f3f3f;
const double tol = 1e-8;
const int N = 1000005;
int primes[N], minp[N], a[N], left_pos[N];
int liner_sieve(int n, int primes[], int minp[]) {
  int cnt = 0;
  memset(minp, 0, n * sizeof *minp);
  for (int i = 2; i < n; ++i) {
    if (!minp[i]) {
      minp[i] = primes[cnt++] = i;
    }
    for (int j = 0; primes[j] * i < n; ++j) {
      minp[primes[j] * i] = primes[j];
      if (i % primes[j] == 0) break;
    }
  }
  return cnt;
}
vector<int> pos[N];
void add(int idx, int num) {
  left_pos[idx] = idx;
  for (int p = minp[num]; num != 1; p = minp[num /= p]) pos[p].push_back(idx);
}
void del(int idx, int num) {
  left_pos[idx] = idx;
  for (int p = minp[num]; num != 1; p = minp[num /= p]) {
    if (pos[p].size() == 0) {
      left_pos[idx] = 0;
      return;
    }
    minx(left_pos[idx], pos[p].back());
    pos[p].pop_back();
  }
}
int main() {
  int n;
  cin >> n;
  liner_sieve(N, primes, minp);
  for (int i = 1; i <= n; ++i) a[i] = int_in();
  char type;
  for (int i = 1; i <= n; ++i) {
    scanf("%c", &type);
    if (type == '*')
      add(i, a[i]);
    else
      del(i, a[i]);
  }
  vector<int> stk;
  left_pos[n + 1] = -1;
  stk.push_back(n + 1);
  long long ans = 0;
  for (int i = n; i >= 1; --i) {
    while (left_pos[stk.back()] >= left_pos[i]) {
      stk.pop_back();
    }
    if (left_pos[i] == i) ans += stk.back() - i;
    stk.push_back(i);
  }
  printf("%lld\n", ans);
  return 0;
}

```