#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n, a[N], pri[N], f[N], stk[N], top, lst[N];
char s[N];
long long ans;
void init() {
  for (int i = 2; i < N; ++i) {
    if (!pri[i]) {
      pri[i] = i;
      if (i > 1000) continue;
      for (int j = i * i; j < N; j += i) pri[j] = i;
    }
  }
}
vector<int> st[N];
void insert(int pos, int val) {
  lst[pos] = pos;
  while (val > 1) {
    st[pri[val]].push_back(pos);
    val = val / pri[val];
  }
}
void del(int pos, int val) {
  if (val == 1) {
    lst[pos] = pos;
    return;
  }
  int Min = pos;
  while (val > 1) {
    int tmp = pri[val];
    if (st[tmp].size() == 0) {
      lst[pos] = 0;
      return;
    }
    Min = min(Min, st[tmp].back());
    st[tmp].pop_back();
    val = val / tmp;
  }
  lst[pos] = Min;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  cin >> s + 1;
  init();
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '*')
      insert(i, a[i]);
    else
      del(i, a[i]);
  }
  for (int i = n; i >= 1; --i) {
    int cnt = 1;
    while (top && lst[stk[top]] >= lst[i]) {
      cnt += f[stk[top]];
      top--;
    }
    f[i] = cnt, stk[++top] = i;
    if (lst[i] == i) ans += f[i];
  }
  cout << ans << endl;
  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 N = 1e6 + 10;
int n, a[N], pri[N], f[N], stk[N], top, lst[N];
char s[N];
long long ans;
void init() {
  for (int i = 2; i < N; ++i) {
    if (!pri[i]) {
      pri[i] = i;
      if (i > 1000) continue;
      for (int j = i * i; j < N; j += i) pri[j] = i;
    }
  }
}
vector<int> st[N];
void insert(int pos, int val) {
  lst[pos] = pos;
  while (val > 1) {
    st[pri[val]].push_back(pos);
    val = val / pri[val];
  }
}
void del(int pos, int val) {
  if (val == 1) {
    lst[pos] = pos;
    return;
  }
  int Min = pos;
  while (val > 1) {
    int tmp = pri[val];
    if (st[tmp].size() == 0) {
      lst[pos] = 0;
      return;
    }
    Min = min(Min, st[tmp].back());
    st[tmp].pop_back();
    val = val / tmp;
  }
  lst[pos] = Min;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  cin >> s + 1;
  init();
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '*')
      insert(i, a[i]);
    else
      del(i, a[i]);
  }
  for (int i = n; i >= 1; --i) {
    int cnt = 1;
    while (top && lst[stk[top]] >= lst[i]) {
      cnt += f[stk[top]];
      top--;
    }
    f[i] = cnt, stk[++top] = i;
    if (lst[i] == i) ans += f[i];
  }
  cout << ans << endl;
  return 0;
}

```