#include <bits/stdc++.h>
using namespace std;
const int M = 1000005;
int read() {
  int x = 0, f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9') {
    if (c == '-') f = -1;
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
int n, m, a[M], b[M], p[M], s[M], num[M];
vector<int> v[M];
char t[M];
long long ans;
void add(int x, int i) {
  b[i] = i;
  while (x > 1) {
    v[p[x]].push_back(i);
    x /= p[x];
  }
}
void del(int x, int i) {
  b[i] = i;
  while (x > 1) {
    if (v[p[x]].empty()) {
      b[i] = 0;
      return;
    }
    b[i] = min(b[i], v[p[x]].back());
    v[p[x]].pop_back();
    x /= p[x];
  }
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 2; i <= 1e6; i++)
    if (!p[i]) {
      p[i] = i;
      for (int j = i; j <= 1e6; j += i) p[j] = i;
    }
  scanf("%s", t + 1);
  for (int i = 1; i <= n; i++) {
    if (t[i] == '*')
      add(a[i], i);
    else
      del(a[i], i);
  }
  for (int i = n; i >= 1; i--) {
    int cnt = 1;
    while (m > 0 && s[m] >= b[i]) {
      cnt += num[m];
      m--;
    }
    s[++m] = b[i];
    num[m] = cnt;
    if (i == b[i]) ans += cnt;
  }
  printf("%lld\n", ans);
}

### 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;
const int M = 1000005;
int read() {
  int x = 0, f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9') {
    if (c == '-') f = -1;
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
int n, m, a[M], b[M], p[M], s[M], num[M];
vector<int> v[M];
char t[M];
long long ans;
void add(int x, int i) {
  b[i] = i;
  while (x > 1) {
    v[p[x]].push_back(i);
    x /= p[x];
  }
}
void del(int x, int i) {
  b[i] = i;
  while (x > 1) {
    if (v[p[x]].empty()) {
      b[i] = 0;
      return;
    }
    b[i] = min(b[i], v[p[x]].back());
    v[p[x]].pop_back();
    x /= p[x];
  }
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 2; i <= 1e6; i++)
    if (!p[i]) {
      p[i] = i;
      for (int j = i; j <= 1e6; j += i) p[j] = i;
    }
  scanf("%s", t + 1);
  for (int i = 1; i <= n; i++) {
    if (t[i] == '*')
      add(a[i], i);
    else
      del(a[i], i);
  }
  for (int i = n; i >= 1; i--) {
    int cnt = 1;
    while (m > 0 && s[m] >= b[i]) {
      cnt += num[m];
      m--;
    }
    s[++m] = b[i];
    num[m] = cnt;
    if (i == b[i]) ans += cnt;
  }
  printf("%lld\n", ans);
}

```