#include <bits/stdc++.h>
int n, a[1000005], from[1000005];
long long ans;
char b[1000005];
std::vector<int> p;
std::vector<std::pair<int, int>> app[1000005], sta, w[1000005];
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin >> n;
  for (int i = 1; i <= n; ++i) std::cin >> a[i];
  for (int i = 1; i <= n; ++i) std::cin >> b[i];
  for (int i = 2; i <= 1000000; ++i) {
    if (!from[i]) {
      from[i] = i;
      p.push_back(i);
    }
    for (auto j : p) {
      if (i * j > 1000000) break;
      from[i * j] = j;
      if (i % j == 0) break;
    }
  }
  for (int i = 1; i <= n; ++i)
    while (a[i] > 1) {
      int v = from[a[i]], tim = 0;
      while (a[i] % v == 0) a[i] /= v, ++tim;
      if (app[v].empty()) app[v].emplace_back(0, 0);
      app[v].push_back({(b[i] == '/' ? -1 : +1) * tim, i});
    }
  for (int i = 1; i <= 1000000; ++i) {
    if (app[i].empty()) continue;
    for (int j = 1; j != (int)app[i].size(); ++j)
      app[i][j].first += app[i][j - 1].first;
    app[i].push_back({INT_MIN, n + 1});
    sta.clear();
    sta.emplace_back(INT_MIN, n + 1);
    for (int j = app[i].size() - 1; j--;) {
      while (sta.back().first >= app[i][j].first) sta.pop_back();
      w[app[i][j + 1].second].emplace_back(sta.back().second, app[i][j].second);
      sta.emplace_back(app[i][j]);
    }
  }
  std::priority_queue<std::pair<int, int>, std::vector<std::pair<int, int>>,
                      std::greater<std::pair<int, int>>>
      que;
  que.push({n + 1, 0});
  for (int i = n; i; --i) {
    for (auto j : w[i]) que.push(j);
    while (que.top().second >= i) que.pop();
    ans += que.top().first - i;
  }
  std::cout << ans << std::endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
int n, a[1000005], from[1000005];
long long ans;
char b[1000005];
std::vector<int> p;
std::vector<std::pair<int, int>> app[1000005], sta, w[1000005];
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin >> n;
  for (int i = 1; i <= n; ++i) std::cin >> a[i];
  for (int i = 1; i <= n; ++i) std::cin >> b[i];
  for (int i = 2; i <= 1000000; ++i) {
    if (!from[i]) {
      from[i] = i;
      p.push_back(i);
    }
    for (auto j : p) {
      if (i * j > 1000000) break;
      from[i * j] = j;
      if (i % j == 0) break;
    }
  }
  for (int i = 1; i <= n; ++i)
    while (a[i] > 1) {
      int v = from[a[i]], tim = 0;
      while (a[i] % v == 0) a[i] /= v, ++tim;
      if (app[v].empty()) app[v].emplace_back(0, 0);
      app[v].push_back({(b[i] == '/' ? -1 : +1) * tim, i});
    }
  for (int i = 1; i <= 1000000; ++i) {
    if (app[i].empty()) continue;
    for (int j = 1; j != (int)app[i].size(); ++j)
      app[i][j].first += app[i][j - 1].first;
    app[i].push_back({INT_MIN, n + 1});
    sta.clear();
    sta.emplace_back(INT_MIN, n + 1);
    for (int j = app[i].size() - 1; j--;) {
      while (sta.back().first >= app[i][j].first) sta.pop_back();
      w[app[i][j + 1].second].emplace_back(sta.back().second, app[i][j].second);
      sta.emplace_back(app[i][j]);
    }
  }
  std::priority_queue<std::pair<int, int>, std::vector<std::pair<int, int>>,
                      std::greater<std::pair<int, int>>>
      que;
  que.push({n + 1, 0});
  for (int i = n; i; --i) {
    for (auto j : w[i]) que.push(j);
    while (que.top().second >= i) que.pop();
    ans += que.top().first - i;
  }
  std::cout << ans << std::endl;
  return 0;
}

```