#include <bits/stdc++.h>
using namespace std;
using LL = long long;
constexpr int maxn = 1000000 + 1;
vector<pair<int, int>> vp[maxn], ch[maxn];
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  for (int i = 2; i < maxn; i += 1)
    if (vp[i].empty())
      for (int j = i; j < maxn; j += i) {
        vp[j].emplace_back(i, 0);
        for (int k = j; k % i == 0; k /= i) vp[j].back().second += 1;
      }
  int n;
  cin >> n;
  vector<int> a(n);
  for (int& x : a) cin >> x;
  string s;
  cin >> s;
  vector<vector<int>> L(n), R(n);
  for (int i = 0; i < n; i += 1)
    for (auto [p, c] : vp[a[i]]) ch[p].emplace_back(i, s[i] == '*' ? c : -c);
  for (int i = 2; i < maxn; i += 1)
    if (ch[i].size()) {
      int m = ch[i].size();
      vector<int> sum(m);
      for (int j = 0; j < m; j += 1) {
        sum[j] = ch[i][j].second;
        if (j) sum[j] += sum[j - 1];
      }
      vector<pair<int, int>> st = {{sum[m - 1], m - 1}};
      for (int j = m - 1; j >= 0; j -= 1) {
        int p = j ? sum[j - 1] : 0;
        while (not st.empty() and st.back().first >= p) st.pop_back();
        st.emplace_back(p, j - 1);
        int tl = (int)st.size() - 2;
        if (tl != -1) {
          int l = j ? ch[i][j - 1].first + 1 : 0;
          int r = ch[i][j].first;
          int x = ch[i][st[tl].second].first - 1;
          L[l].push_back(x);
          R[r].push_back(x);
        }
      }
    }
  LL ans = 0;
  multiset<int> ms;
  for (int i = 0; i < n; i += 1) {
    for (int l : L[i]) ms.insert(l);
    if (ms.empty())
      ans += n - i;
    else
      ans += *ms.begin() - i + 1;
    for (int r : R[i]) ms.erase(ms.find(r));
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
using LL = long long;
constexpr int maxn = 1000000 + 1;
vector<pair<int, int>> vp[maxn], ch[maxn];
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  for (int i = 2; i < maxn; i += 1)
    if (vp[i].empty())
      for (int j = i; j < maxn; j += i) {
        vp[j].emplace_back(i, 0);
        for (int k = j; k % i == 0; k /= i) vp[j].back().second += 1;
      }
  int n;
  cin >> n;
  vector<int> a(n);
  for (int& x : a) cin >> x;
  string s;
  cin >> s;
  vector<vector<int>> L(n), R(n);
  for (int i = 0; i < n; i += 1)
    for (auto [p, c] : vp[a[i]]) ch[p].emplace_back(i, s[i] == '*' ? c : -c);
  for (int i = 2; i < maxn; i += 1)
    if (ch[i].size()) {
      int m = ch[i].size();
      vector<int> sum(m);
      for (int j = 0; j < m; j += 1) {
        sum[j] = ch[i][j].second;
        if (j) sum[j] += sum[j - 1];
      }
      vector<pair<int, int>> st = {{sum[m - 1], m - 1}};
      for (int j = m - 1; j >= 0; j -= 1) {
        int p = j ? sum[j - 1] : 0;
        while (not st.empty() and st.back().first >= p) st.pop_back();
        st.emplace_back(p, j - 1);
        int tl = (int)st.size() - 2;
        if (tl != -1) {
          int l = j ? ch[i][j - 1].first + 1 : 0;
          int r = ch[i][j].first;
          int x = ch[i][st[tl].second].first - 1;
          L[l].push_back(x);
          R[r].push_back(x);
        }
      }
    }
  LL ans = 0;
  multiset<int> ms;
  for (int i = 0; i < n; i += 1) {
    for (int l : L[i]) ms.insert(l);
    if (ms.empty())
      ans += n - i;
    else
      ans += *ms.begin() - i + 1;
    for (int r : R[i]) ms.erase(ms.find(r));
  }
  cout << ans << "\n";
  return 0;
}

```