#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long INF = 1e18 + 123;
const long double EPS = 1e-9;
const int inf = 1e9 + 123;
const int MOD = 1e9 + 7;
const int N = 2e5 + 123;
const int M = 1e6 + 12;
const double pi = 3.14159265359;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
int a[M];
vector<pair<int, int> > v[M];
bool bp[1000];
void solve() {
  vector<int> pr;
  for (int i = 2; i < 1000; i++) {
    if (bp[i]) continue;
    pr.push_back(i);
    for (int j = i * i; j < 1000; j += i) {
      bp[j] = 1;
    }
  }
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  string s;
  cin >> s;
  long long ans = 0;
  set<int> bb = {inf};
  for (int i = n, mn; i >= 1; i--) {
    mn = n + 1;
    if (a[i] > 1 && s[i - 1] == '/') mn = i;
    for (int d : pr) {
      if (a[i] % d) {
        if (!v[d].empty()) {
          mn = min(mn, v[d].back().second);
        }
        continue;
      }
      int cnt = 0;
      while (a[i] % d == 0) {
        a[i] /= d;
        cnt++;
      }
      if (s[i - 1] == '*') {
        while (!v[d].empty()) {
          if (v[d].back().first <= cnt) {
            cnt -= v[d].back().first;
            v[d].pop_back();
          } else {
            mn = min(mn, v[d].back().second);
            (--v[d].end())->first -= cnt;
            break;
          }
        }
      } else {
        v[d].push_back({cnt, i});
      }
    }
    if (a[i] > 1) {
      int d = a[i];
      if (s[i - 1] == '*') {
        if (!v[d].empty()) {
          bb.erase(v[d].back().second);
          v[d].pop_back();
        }
      } else {
        v[d].push_back({1, i});
        bb.insert(i);
      }
    }
    mn = min(mn, *bb.begin());
    ans += mn - i;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long INF = 1e18 + 123;
const long double EPS = 1e-9;
const int inf = 1e9 + 123;
const int MOD = 1e9 + 7;
const int N = 2e5 + 123;
const int M = 1e6 + 12;
const double pi = 3.14159265359;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
int a[M];
vector<pair<int, int> > v[M];
bool bp[1000];
void solve() {
  vector<int> pr;
  for (int i = 2; i < 1000; i++) {
    if (bp[i]) continue;
    pr.push_back(i);
    for (int j = i * i; j < 1000; j += i) {
      bp[j] = 1;
    }
  }
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  string s;
  cin >> s;
  long long ans = 0;
  set<int> bb = {inf};
  for (int i = n, mn; i >= 1; i--) {
    mn = n + 1;
    if (a[i] > 1 && s[i - 1] == '/') mn = i;
    for (int d : pr) {
      if (a[i] % d) {
        if (!v[d].empty()) {
          mn = min(mn, v[d].back().second);
        }
        continue;
      }
      int cnt = 0;
      while (a[i] % d == 0) {
        a[i] /= d;
        cnt++;
      }
      if (s[i - 1] == '*') {
        while (!v[d].empty()) {
          if (v[d].back().first <= cnt) {
            cnt -= v[d].back().first;
            v[d].pop_back();
          } else {
            mn = min(mn, v[d].back().second);
            (--v[d].end())->first -= cnt;
            break;
          }
        }
      } else {
        v[d].push_back({cnt, i});
      }
    }
    if (a[i] > 1) {
      int d = a[i];
      if (s[i - 1] == '*') {
        if (!v[d].empty()) {
          bb.erase(v[d].back().second);
          v[d].pop_back();
        }
      } else {
        v[d].push_back({1, i});
        bb.insert(i);
      }
    }
    mn = min(mn, *bb.begin());
    ans += mn - i;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```