#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0;
  bool flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return flag ? x : ~(x - 1);
}
inline void write(long long x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const long long N = 1e6 + 10;
long long n, ans, a[N];
long long prime[N], mp[N], cnt;
long long stk[N], top, pre[N], w[N];
bool is_prime[N];
vector<long long> vec[N];
inline void sieve(long long x) {
  for (long long i = 2; i <= x; i++) {
    if (!is_prime[i]) prime[++cnt] = mp[i] = i;
    for (long long j = 1; prime[j] <= x / i; j++) {
      is_prime[i * prime[j]] = true;
      mp[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
signed main() {
  sieve(1000000);
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i <= n; i++) {
    char ch = getchar();
    while (ch != '*' && ch != '/') ch = getchar();
    if (ch == '*') {
      while (a[i] > 1) {
        long long p = mp[a[i]];
        vec[p].push_back(i);
        a[i] /= p;
      }
      pre[i] = i;
    } else {
      pre[i] = i;
      while (a[i] > 1) {
        long long p = mp[a[i]];
        if (vec[p].empty())
          pre[i] = -1;
        else
          pre[i] = min(pre[i], vec[p].back()), vec[p].pop_back();
        a[i] /= p;
      }
    }
  }
  for (long long i = n; i >= 1; i--) {
    long long nw = 1;
    while (top && pre[i] <= stk[top]) nw += w[top], top--;
    stk[++top] = pre[i], w[top] = nw;
    if (i == pre[i]) ans += nw;
  }
  write(ans), putchar('\n');
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
inline long long read() {
  long long x = 0;
  bool flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return flag ? x : ~(x - 1);
}
inline void write(long long x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const long long N = 1e6 + 10;
long long n, ans, a[N];
long long prime[N], mp[N], cnt;
long long stk[N], top, pre[N], w[N];
bool is_prime[N];
vector<long long> vec[N];
inline void sieve(long long x) {
  for (long long i = 2; i <= x; i++) {
    if (!is_prime[i]) prime[++cnt] = mp[i] = i;
    for (long long j = 1; prime[j] <= x / i; j++) {
      is_prime[i * prime[j]] = true;
      mp[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
signed main() {
  sieve(1000000);
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i <= n; i++) {
    char ch = getchar();
    while (ch != '*' && ch != '/') ch = getchar();
    if (ch == '*') {
      while (a[i] > 1) {
        long long p = mp[a[i]];
        vec[p].push_back(i);
        a[i] /= p;
      }
      pre[i] = i;
    } else {
      pre[i] = i;
      while (a[i] > 1) {
        long long p = mp[a[i]];
        if (vec[p].empty())
          pre[i] = -1;
        else
          pre[i] = min(pre[i], vec[p].back()), vec[p].pop_back();
        a[i] /= p;
      }
    }
  }
  for (long long i = n; i >= 1; i--) {
    long long nw = 1;
    while (top && pre[i] <= stk[top]) nw += w[top], top--;
    stk[++top] = pre[i], w[top] = nw;
    if (i == pre[i]) ans += nw;
  }
  write(ans), putchar('\n');
  return 0;
}

```