#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
int a[maxn], prim[maxn], tot, flag[maxn], mi[maxn], l[maxn];
vector<int> st[maxn];
void init(int n) {
  mi[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!flag[i]) prim[tot++] = i, mi[i] = i;
    for (int j = 0; j < tot && prim[j] * i <= n; ++j) {
      flag[i * prim[j]] = 1;
      if (i % prim[j] == 0) {
        mi[i * prim[j]] = mi[i];
        break;
      }
      mi[i * prim[j]] = prim[j];
    }
  }
}
char s[maxn];
int main() {
  init(maxn - 1);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  scanf("%s", s + 1);
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '*') {
      l[i] = i;
      int u = a[i];
      while (u != 1) {
        st[mi[u]].push_back(i);
        u /= mi[u];
      }
    } else {
      l[i] = i;
      int u = a[i];
      while (u != 1) {
        if (st[mi[u]].empty()) {
          l[i] = 0;
        } else {
          l[i] = min(l[i], st[mi[u]].back());
          st[mi[u]].pop_back();
        }
        u /= mi[u];
      }
    }
  }
  vector<pair<int, int>> st2;
  long long ans = 0;
  for (int i = n; i >= 1; --i) {
    int cnt = 1;
    while (!st2.empty() && st2.back().first >= i) {
      cnt += st2.back().second;
      st2.pop_back();
    }
    st2.push_back({l[i], cnt});
    if (l[i] == i) ans += cnt;
  }
  printf("%lld\n", ans);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 maxn = 1e6 + 5;
int a[maxn], prim[maxn], tot, flag[maxn], mi[maxn], l[maxn];
vector<int> st[maxn];
void init(int n) {
  mi[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!flag[i]) prim[tot++] = i, mi[i] = i;
    for (int j = 0; j < tot && prim[j] * i <= n; ++j) {
      flag[i * prim[j]] = 1;
      if (i % prim[j] == 0) {
        mi[i * prim[j]] = mi[i];
        break;
      }
      mi[i * prim[j]] = prim[j];
    }
  }
}
char s[maxn];
int main() {
  init(maxn - 1);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  scanf("%s", s + 1);
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '*') {
      l[i] = i;
      int u = a[i];
      while (u != 1) {
        st[mi[u]].push_back(i);
        u /= mi[u];
      }
    } else {
      l[i] = i;
      int u = a[i];
      while (u != 1) {
        if (st[mi[u]].empty()) {
          l[i] = 0;
        } else {
          l[i] = min(l[i], st[mi[u]].back());
          st[mi[u]].pop_back();
        }
        u /= mi[u];
      }
    }
  }
  vector<pair<int, int>> st2;
  long long ans = 0;
  for (int i = n; i >= 1; --i) {
    int cnt = 1;
    while (!st2.empty() && st2.back().first >= i) {
      cnt += st2.back().second;
      st2.pop_back();
    }
    st2.push_back({l[i], cnt});
    if (l[i] == i) ans += cnt;
  }
  printf("%lld\n", ans);
}

```