#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000005;
int n, top, mx;
long long ans;
int a[maxn], stk[maxn];
vector<int> p[maxn], v[maxn];
string s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), mx = max(mx, a[i]);
  for (int i = 2; i <= mx; i++)
    if (v[i].empty())
      for (int j = i; j <= mx; j += i) {
        int k = j;
        while (k % i == 0) k /= i, v[j].push_back(i);
      }
  cin >> s, s = " " + s;
  for (int i = 1; i <= n; i++) {
    stk[++top] = i;
    if (s[i] == '*')
      for (int j = 0; j < v[a[i]].size(); j++) p[v[a[i]][j]].push_back(i);
    if (s[i] == '/')
      for (int j = 0; j < v[a[i]].size(); j++) {
        if (p[v[a[i]][j]].size() == 0) {
          top = 0;
          break;
        }
        while (top && stk[top] > p[v[a[i]][j]].back()) top--;
        p[v[a[i]][j]].pop_back();
      }
    ans += top;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 maxn = 1000005;
int n, top, mx;
long long ans;
int a[maxn], stk[maxn];
vector<int> p[maxn], v[maxn];
string s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), mx = max(mx, a[i]);
  for (int i = 2; i <= mx; i++)
    if (v[i].empty())
      for (int j = i; j <= mx; j += i) {
        int k = j;
        while (k % i == 0) k /= i, v[j].push_back(i);
      }
  cin >> s, s = " " + s;
  for (int i = 1; i <= n; i++) {
    stk[++top] = i;
    if (s[i] == '*')
      for (int j = 0; j < v[a[i]].size(); j++) p[v[a[i]][j]].push_back(i);
    if (s[i] == '/')
      for (int j = 0; j < v[a[i]].size(); j++) {
        if (p[v[a[i]][j]].size() == 0) {
          top = 0;
          break;
        }
        while (top && stk[top] > p[v[a[i]][j]].back()) top--;
        p[v[a[i]][j]].pop_back();
      }
    ans += top;
  }
  printf("%lld\n", ans);
  return 0;
}

```