#include <bits/stdc++.h>
using namespace std;
int n;
string s;
int a[1000001];
int b[1000001];
int p[1000001];
int f[1000001];
vector<pair<int, int>> pos[1000001];
class segment_tree {
 public:
  using pointer = segment_tree*;
  pointer left, right;
  int l, r, m;
  int val;
  segment_tree(int l, int r)
      : l(l), r(r), m((l + r) / 2), val(n), left(nullptr), right(nullptr) {
    if (l != r) {
      left = new segment_tree(l, m);
      right = new segment_tree(m + 1, r);
    }
  }
  void update(int u, int v, int a) {
    if (l > v || r < u) return;
    if (u <= l && v >= r)
      val = min(val, a);
    else {
      left->update(u, v, a);
      right->update(u, v, a);
    }
  }
  void get(int best = n) {
    best = min(best, val);
    if (l == r)
      f[l] = best;
    else {
      left->get(best);
      right->get(best);
    }
  }
};
segment_tree* tree;
void add_query(int x, int y, int p) { tree->update(x, y, p); }
void make(vector<pair<int, int>>& pos) {
  for (int i = 1; i < pos.size(); i++) pos[i].second += pos[i - 1].second;
  stack<int> temp;
  for (int i = ((int)pos.size()) - 1; i >= 0; i--) {
    while ((!temp.empty()) && (pos[i].second <= pos[temp.top()].second))
      temp.pop();
    int best = temp.empty() ? (n + 1) : pos[temp.top()].first;
    if (i + 1 >= pos.size())
      add_query(pos[i].first + 1, n, best - 1);
    else
      add_query(pos[i].first + 1, pos[i + 1].first, best - 1);
    temp.push(i);
  }
  while ((!temp.empty()) && (0 <= pos[temp.top()].second)) temp.pop();
  int best = temp.empty() ? (n + 1) : pos[temp.top()].first;
  add_query(1, pos[0].first, best - 1);
}
void solve() {
  for (int i = 2; i <= 1000000; i++)
    if (p[i] == 0)
      for (int j = i; j <= 1000000; j += i)
        if (p[j] == 0) p[j] = i;
  cin >> n;
  tree = new segment_tree(1, n);
  for (int i = 1; i <= n; i++) cin >> a[i];
  cin >> s;
  for (int i = 1; i <= n; i++)
    if (s[i - 1] == '*')
      b[i] = 1;
    else
      b[i] = -1;
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    while (x > 1) {
      int k = p[x];
      pos[k].emplace_back(i, 0);
      while (x % k == 0) {
        pos[k].back().second += b[i];
        x /= k;
      }
    }
  }
  for (int i = 1; i <= 1000000; i++)
    if (!pos[i].empty()) {
      make(pos[i]);
    }
  tree->get();
  int64_t ans = 0;
  for (int i = 1; i <= n; i++) ans += (f[i] - i + 1);
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Generate 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;
int n;
string s;
int a[1000001];
int b[1000001];
int p[1000001];
int f[1000001];
vector<pair<int, int>> pos[1000001];
class segment_tree {
 public:
  using pointer = segment_tree*;
  pointer left, right;
  int l, r, m;
  int val;
  segment_tree(int l, int r)
      : l(l), r(r), m((l + r) / 2), val(n), left(nullptr), right(nullptr) {
    if (l != r) {
      left = new segment_tree(l, m);
      right = new segment_tree(m + 1, r);
    }
  }
  void update(int u, int v, int a) {
    if (l > v || r < u) return;
    if (u <= l && v >= r)
      val = min(val, a);
    else {
      left->update(u, v, a);
      right->update(u, v, a);
    }
  }
  void get(int best = n) {
    best = min(best, val);
    if (l == r)
      f[l] = best;
    else {
      left->get(best);
      right->get(best);
    }
  }
};
segment_tree* tree;
void add_query(int x, int y, int p) { tree->update(x, y, p); }
void make(vector<pair<int, int>>& pos) {
  for (int i = 1; i < pos.size(); i++) pos[i].second += pos[i - 1].second;
  stack<int> temp;
  for (int i = ((int)pos.size()) - 1; i >= 0; i--) {
    while ((!temp.empty()) && (pos[i].second <= pos[temp.top()].second))
      temp.pop();
    int best = temp.empty() ? (n + 1) : pos[temp.top()].first;
    if (i + 1 >= pos.size())
      add_query(pos[i].first + 1, n, best - 1);
    else
      add_query(pos[i].first + 1, pos[i + 1].first, best - 1);
    temp.push(i);
  }
  while ((!temp.empty()) && (0 <= pos[temp.top()].second)) temp.pop();
  int best = temp.empty() ? (n + 1) : pos[temp.top()].first;
  add_query(1, pos[0].first, best - 1);
}
void solve() {
  for (int i = 2; i <= 1000000; i++)
    if (p[i] == 0)
      for (int j = i; j <= 1000000; j += i)
        if (p[j] == 0) p[j] = i;
  cin >> n;
  tree = new segment_tree(1, n);
  for (int i = 1; i <= n; i++) cin >> a[i];
  cin >> s;
  for (int i = 1; i <= n; i++)
    if (s[i - 1] == '*')
      b[i] = 1;
    else
      b[i] = -1;
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    while (x > 1) {
      int k = p[x];
      pos[k].emplace_back(i, 0);
      while (x % k == 0) {
        pos[k].back().second += b[i];
        x /= k;
      }
    }
  }
  for (int i = 1; i <= 1000000; i++)
    if (!pos[i].empty()) {
      make(pos[i]);
    }
  tree->get();
  int64_t ans = 0;
  for (int i = 1; i <= n; i++) ans += (f[i] - i + 1);
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

```