#include <bits/stdc++.h>
using namespace std;
const int N = 1000000;
int n, a[N + 9];
char s[N + 9];
void into() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  scanf("%s", s + 1);
}
struct factor {
  int x, c;
  factor(int X = 0, int C = 0) : x(X), c(C) {}
};
vector<factor> d[N + 9];
void Get_d() {
  for (int i = 2; i <= N; ++i) {
    if (!d[i].empty()) continue;
    for (int j = i; j <= N; j += i) {
      int t = j, c = 0;
      for (; t % i == 0; t /= i) ++c;
      d[j].emplace_back(i, c);
    }
  }
}
int uni[N + 9];
int Query_uni(int k) { return k == uni[k] ? k : uni[k] = Query_uni(uni[k]); }
struct state {
  int l, r, v;
  state(int L = 0, int R = 0, int V = 0) : l(L), r(R), v(V) {}
  bool operator<(const state &p) const { return v < p.v; }
};
priority_queue<state> q[N + 9];
int last[N + 9], cnt[N + 9];
long long ans;
void Get_ans() {
  for (int i = 0; i <= n; ++i) uni[i] = i;
  for (int i = 1, now = 0; i <= n; ++i) {
    ++now;
    for (auto [x, c] : d[a[i]]) {
      q[x].emplace(last[x] + 1, i, cnt[x]);
      cnt[x] += (s[i] == '*' ? 1 : -1) * c;
      last[x] = i;
      for (; !q[x].empty() && q[x].top().v > cnt[x];) {
        auto [l, r, rub] = q[x].top();
        q[x].pop();
        for (int i = Query_uni(r); i >= l; uni[i] = i - 1, i = Query_uni(i))
          --now;
      }
    }
    if (s[i] == '/' && a[i] > 1 && uni[i] == i) uni[i] = i - 1, --now;
    ans += now;
  }
}
void work() {
  Get_d();
  Get_ans();
}
void outo() { printf("%lld\n", ans); }
int main() {
  int T = 1;
  for (; T--;) {
    into();
    work();
    outo();
  }
  return 0;
}

### 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;
const int N = 1000000;
int n, a[N + 9];
char s[N + 9];
void into() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  scanf("%s", s + 1);
}
struct factor {
  int x, c;
  factor(int X = 0, int C = 0) : x(X), c(C) {}
};
vector<factor> d[N + 9];
void Get_d() {
  for (int i = 2; i <= N; ++i) {
    if (!d[i].empty()) continue;
    for (int j = i; j <= N; j += i) {
      int t = j, c = 0;
      for (; t % i == 0; t /= i) ++c;
      d[j].emplace_back(i, c);
    }
  }
}
int uni[N + 9];
int Query_uni(int k) { return k == uni[k] ? k : uni[k] = Query_uni(uni[k]); }
struct state {
  int l, r, v;
  state(int L = 0, int R = 0, int V = 0) : l(L), r(R), v(V) {}
  bool operator<(const state &p) const { return v < p.v; }
};
priority_queue<state> q[N + 9];
int last[N + 9], cnt[N + 9];
long long ans;
void Get_ans() {
  for (int i = 0; i <= n; ++i) uni[i] = i;
  for (int i = 1, now = 0; i <= n; ++i) {
    ++now;
    for (auto [x, c] : d[a[i]]) {
      q[x].emplace(last[x] + 1, i, cnt[x]);
      cnt[x] += (s[i] == '*' ? 1 : -1) * c;
      last[x] = i;
      for (; !q[x].empty() && q[x].top().v > cnt[x];) {
        auto [l, r, rub] = q[x].top();
        q[x].pop();
        for (int i = Query_uni(r); i >= l; uni[i] = i - 1, i = Query_uni(i))
          --now;
      }
    }
    if (s[i] == '/' && a[i] > 1 && uni[i] == i) uni[i] = i - 1, --now;
    ans += now;
  }
}
void work() {
  Get_d();
  Get_ans();
}
void outo() { printf("%lld\n", ans); }
int main() {
  int T = 1;
  for (; T--;) {
    into();
    work();
    outo();
  }
  return 0;
}

```