#include <bits/stdc++.h>
using namespace std;
const int INF = 100000000;
const int MAX = 1000000;
template <typename T>
struct dual_segment_tree {
  int N;
  vector<T> ST;
  dual_segment_tree(int n, int x) {
    N = 1;
    while (N < n) {
      N *= 2;
    }
    ST = vector<T>(N * 2 - 1, x);
  }
  T operator[](int k) {
    k += N - 1;
    T ans = ST[k];
    while (k > 0) {
      k = (k - 1) / 2;
      ans = min(ans, ST[k]);
    }
    return ans;
  }
  void range_chmin(int L, int R, T x, int i, int l, int r) {
    if (R <= l || r <= L) {
      return;
    } else if (L <= l && r <= R) {
      ST[i] = min(ST[i], x);
      return;
    } else {
      int m = (l + r) / 2;
      range_chmin(L, R, x, i * 2 + 1, l, m);
      range_chmin(L, R, x, i * 2 + 2, m, r);
      return;
    }
  }
  void range_chmin(int L, int R, T x) { range_chmin(L, R, x, 0, 0, N); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<char> b(n);
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  vector<int> prime(MAX + 1, -1);
  for (int i = 2; i <= MAX; i++) {
    if (prime[i] == -1) {
      for (int j = i * 2; j <= MAX; j += i) {
        prime[j] = i;
      }
    }
  }
  vector<vector<pair<int, int>>> id(MAX);
  for (int i = 0; i < n; i++) {
    int x = a[i];
    while (prime[x] != -1) {
      int p = prime[x];
      int cnt = 0;
      while (x % p == 0) {
        x /= p;
        cnt++;
      }
      if (b[i] == '*') {
        id[p].push_back(make_pair(cnt, i));
      } else {
        id[p].push_back(make_pair(-cnt, i));
      }
    }
    if (x > 1) {
      if (b[i] == '*') {
        id[x].push_back(make_pair(1, i));
      } else {
        id[x].push_back(make_pair(-1, i));
      }
    }
  }
  dual_segment_tree<int> ST(n, n);
  for (int i = 0; i < MAX; i++) {
    if (!id[i].empty()) {
      int cnt = id[i].size();
      stack<pair<int, int>> st;
      st.push(make_pair(-INF, n + 1));
      st.push(make_pair(0, n));
      for (int j = cnt - 1; j >= 0; j--) {
        int p = id[i][j].second;
        int pp;
        if (j == 0) {
          pp = -1;
        } else {
          pp = id[i][j - 1].second;
        }
        if (id[i][j].first < 0) {
          ST.range_chmin(pp + 1, p + 1, p);
          int nxt = st.top().first - id[i][j].first;
          st.push(make_pair(nxt, p));
        } else {
          int nxt = st.top().first - id[i][j].first;
          while (st.size() >= 2) {
            pair<int, int> P = st.top();
            st.pop();
            if (st.top().first < nxt) {
              st.push(P);
              break;
            }
          }
          int r = st.top().second;
          ST.range_chmin(pp + 1, p + 1, r);
          st.push(make_pair(nxt, p));
        }
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += ST[i] - i;
  }
  cout << ans << endl;
}

### 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 INF = 100000000;
const int MAX = 1000000;
template <typename T>
struct dual_segment_tree {
  int N;
  vector<T> ST;
  dual_segment_tree(int n, int x) {
    N = 1;
    while (N < n) {
      N *= 2;
    }
    ST = vector<T>(N * 2 - 1, x);
  }
  T operator[](int k) {
    k += N - 1;
    T ans = ST[k];
    while (k > 0) {
      k = (k - 1) / 2;
      ans = min(ans, ST[k]);
    }
    return ans;
  }
  void range_chmin(int L, int R, T x, int i, int l, int r) {
    if (R <= l || r <= L) {
      return;
    } else if (L <= l && r <= R) {
      ST[i] = min(ST[i], x);
      return;
    } else {
      int m = (l + r) / 2;
      range_chmin(L, R, x, i * 2 + 1, l, m);
      range_chmin(L, R, x, i * 2 + 2, m, r);
      return;
    }
  }
  void range_chmin(int L, int R, T x) { range_chmin(L, R, x, 0, 0, N); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<char> b(n);
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  vector<int> prime(MAX + 1, -1);
  for (int i = 2; i <= MAX; i++) {
    if (prime[i] == -1) {
      for (int j = i * 2; j <= MAX; j += i) {
        prime[j] = i;
      }
    }
  }
  vector<vector<pair<int, int>>> id(MAX);
  for (int i = 0; i < n; i++) {
    int x = a[i];
    while (prime[x] != -1) {
      int p = prime[x];
      int cnt = 0;
      while (x % p == 0) {
        x /= p;
        cnt++;
      }
      if (b[i] == '*') {
        id[p].push_back(make_pair(cnt, i));
      } else {
        id[p].push_back(make_pair(-cnt, i));
      }
    }
    if (x > 1) {
      if (b[i] == '*') {
        id[x].push_back(make_pair(1, i));
      } else {
        id[x].push_back(make_pair(-1, i));
      }
    }
  }
  dual_segment_tree<int> ST(n, n);
  for (int i = 0; i < MAX; i++) {
    if (!id[i].empty()) {
      int cnt = id[i].size();
      stack<pair<int, int>> st;
      st.push(make_pair(-INF, n + 1));
      st.push(make_pair(0, n));
      for (int j = cnt - 1; j >= 0; j--) {
        int p = id[i][j].second;
        int pp;
        if (j == 0) {
          pp = -1;
        } else {
          pp = id[i][j - 1].second;
        }
        if (id[i][j].first < 0) {
          ST.range_chmin(pp + 1, p + 1, p);
          int nxt = st.top().first - id[i][j].first;
          st.push(make_pair(nxt, p));
        } else {
          int nxt = st.top().first - id[i][j].first;
          while (st.size() >= 2) {
            pair<int, int> P = st.top();
            st.pop();
            if (st.top().first < nxt) {
              st.push(P);
              break;
            }
          }
          int r = st.top().second;
          ST.range_chmin(pp + 1, p + 1, r);
          st.push(make_pair(nxt, p));
        }
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += ST[i] - i;
  }
  cout << ans << endl;
}

```