#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
int f[maxn];
int n;
int a[maxn];
char b[maxn];
int g[maxn];
vector<int> stk[maxn], add[maxn];
void sieve() {
  for (int i = 2; i < maxn; ++i) {
    if (f[i] == 0) {
      for (int j = i; j < maxn; j += i) f[j] = i;
    }
  }
}
struct fenwick {
  int sum[maxn];
  void add(int x) {
    for (; x < maxn; x += x & -x) ++sum[x];
  }
  int get(int x) {
    int res = 0;
    for (; x > 0; x -= x & -x) res += sum[x];
    return res;
  }
  int get(int l, int r) { return get(r) - get(l - 1); }
} tree;
void read_input() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    cin >> b[i];
  }
}
void solve() {
  for (int i = 1; i <= n; ++i) {
    g[i] = i;
    int x = a[i];
    if (b[i] == '*') {
      g[i] = i;
      while (x > 1) {
        stk[f[x]].push_back(i);
        x /= f[x];
      }
    } else {
      while (x > 1) {
        int p = f[x], lst = i;
        bool flag = false;
        while (f[x] == p) {
          if ((int)stk[p].size()) {
            lst = stk[p].back();
            stk[p].pop_back();
          } else
            flag = true;
          x /= p;
        }
        if (flag)
          g[i] = 0;
        else
          g[i] = min(g[i], lst);
      }
    }
    if (g[i]) add[g[i]].push_back(i);
  }
  long long ans = 0;
  for (int i = n; i > 0; --i) {
    for (auto& p : add[i]) tree.add(p);
    int low = i, high = n;
    while (low <= high) {
      int mid = (low + high) / 2;
      if (tree.get(i, mid) == (mid - i + 1))
        low = mid + 1;
      else
        high = mid - 1;
    }
    ans += high - i + 1;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  sieve();
  read_input();
  solve();
}

### 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;
const int maxn = 1e6 + 5;
int f[maxn];
int n;
int a[maxn];
char b[maxn];
int g[maxn];
vector<int> stk[maxn], add[maxn];
void sieve() {
  for (int i = 2; i < maxn; ++i) {
    if (f[i] == 0) {
      for (int j = i; j < maxn; j += i) f[j] = i;
    }
  }
}
struct fenwick {
  int sum[maxn];
  void add(int x) {
    for (; x < maxn; x += x & -x) ++sum[x];
  }
  int get(int x) {
    int res = 0;
    for (; x > 0; x -= x & -x) res += sum[x];
    return res;
  }
  int get(int l, int r) { return get(r) - get(l - 1); }
} tree;
void read_input() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    cin >> b[i];
  }
}
void solve() {
  for (int i = 1; i <= n; ++i) {
    g[i] = i;
    int x = a[i];
    if (b[i] == '*') {
      g[i] = i;
      while (x > 1) {
        stk[f[x]].push_back(i);
        x /= f[x];
      }
    } else {
      while (x > 1) {
        int p = f[x], lst = i;
        bool flag = false;
        while (f[x] == p) {
          if ((int)stk[p].size()) {
            lst = stk[p].back();
            stk[p].pop_back();
          } else
            flag = true;
          x /= p;
        }
        if (flag)
          g[i] = 0;
        else
          g[i] = min(g[i], lst);
      }
    }
    if (g[i]) add[g[i]].push_back(i);
  }
  long long ans = 0;
  for (int i = n; i > 0; --i) {
    for (auto& p : add[i]) tree.add(p);
    int low = i, high = n;
    while (low <= high) {
      int mid = (low + high) / 2;
      if (tree.get(i, mid) == (mid - i + 1))
        low = mid + 1;
      else
        high = mid - 1;
    }
    ans += high - i + 1;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  sieve();
  read_input();
  solve();
}

```