#include <bits/stdc++.h>
using namespace std;
using namespace std;
inline int read() {
  char ch;
  bool sign = 0;
  int res = 0;
  while (!isdigit(ch = getchar()))
    if (ch == '-') sign = 1;
  while (isdigit(ch)) res = res * 10 + (ch ^ 48), ch = getchar();
  if (sign)
    return -res;
  else
    return res;
}
const int maxn = 1000000 + 10;
const int mod = 1e9 + 7;
const int inf = 1e9;
const int maxm = 1000 + 10;
vector<int> prime;
int cnt, vis[maxn];
inline void init() {
  for (int i = 2; i < maxn; ++i) {
    if (!vis[i]) {
      cnt++;
      prime.push_back(i);
    }
    for (int j = 0; j < cnt && prime[j] * i < maxn; ++j) {
      vis[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int a[maxn], n, sum[maxn];
char b[maxn];
int dp[maxn];
inline void brute(int p) {
  for (int i = (1); i <= (n); ++i) {
    int now = 0;
    for (; a[i] % p == 0;) now++, a[i] /= p;
    if (b[i] == '/') now *= -1;
    sum[i] = sum[i - 1] + now;
  }
  sum[n + 1] = -inf;
  vector<int> sk;
  sk.push_back(n + 1);
  for (int i = (n); i >= (0); --i) {
    for (; sum[sk.back()] >= sum[i];) sk.pop_back();
    dp[i + 1] = min(dp[i + 1], sk.back() - 1);
    sk.push_back(i);
  }
}
vector<int> pos[maxn];
int has[maxn];
int f[maxn];
int pre[maxn];
int illpos;
set<int> fval;
inline long long solve() {
  long long ans = 0;
  for (int i = (1); i <= (n); ++i) dp[i] = n;
  for (int i = (0); i < (cnt); ++i) {
    if (prime[i] * prime[i] <= 1000000) {
      brute(prime[i]);
    } else
      break;
  }
  f[n + 1] = n;
  fval.insert(n + 1);
  illpos = n + 1;
  for (int i = (n); i >= (1); --i) {
    assert(vis[a[i]] == 0);
    if (a[i] == 1) {
      f[i] = f[i + 1];
    } else {
      if (b[i] == '/') {
        has[a[i]]--;
        pos[a[i]].push_back(-i);
        fval.erase(pre[a[i]]);
        pre[a[i]] = i;
        fval.insert(i);
        f[i] = i - 1;
      } else {
        if (!pos[a[i]].empty() && pos[a[i]].back() < 0) {
          pos[a[i]].pop_back();
          fval.erase(pre[a[i]]);
        }
        if (!pos[a[i]].empty() && pos[a[i]].back() < 0) {
          pre[a[i]] = -pos[a[i]].back();
          fval.insert(-pos[a[i]].back());
        }
        f[i] = (*fval.begin()) - 1;
      }
    }
  }
  cerr << f[1] << endl;
  for (int i = (1); i <= (n); ++i) {
    assert(min(f[i], dp[i]) + 1 >= i);
    ans += min(f[i], dp[i]) - i + 1;
  }
  return ans;
}
int main(void) {
  init();
  scanf("%d", &(n));
  for (int i = (1); i <= (n); ++i) scanf("%d", &(a[i]));
  scanf("%s", b + 1);
  printf("%lld\n", solve());
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
using namespace std;
inline int read() {
  char ch;
  bool sign = 0;
  int res = 0;
  while (!isdigit(ch = getchar()))
    if (ch == '-') sign = 1;
  while (isdigit(ch)) res = res * 10 + (ch ^ 48), ch = getchar();
  if (sign)
    return -res;
  else
    return res;
}
const int maxn = 1000000 + 10;
const int mod = 1e9 + 7;
const int inf = 1e9;
const int maxm = 1000 + 10;
vector<int> prime;
int cnt, vis[maxn];
inline void init() {
  for (int i = 2; i < maxn; ++i) {
    if (!vis[i]) {
      cnt++;
      prime.push_back(i);
    }
    for (int j = 0; j < cnt && prime[j] * i < maxn; ++j) {
      vis[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int a[maxn], n, sum[maxn];
char b[maxn];
int dp[maxn];
inline void brute(int p) {
  for (int i = (1); i <= (n); ++i) {
    int now = 0;
    for (; a[i] % p == 0;) now++, a[i] /= p;
    if (b[i] == '/') now *= -1;
    sum[i] = sum[i - 1] + now;
  }
  sum[n + 1] = -inf;
  vector<int> sk;
  sk.push_back(n + 1);
  for (int i = (n); i >= (0); --i) {
    for (; sum[sk.back()] >= sum[i];) sk.pop_back();
    dp[i + 1] = min(dp[i + 1], sk.back() - 1);
    sk.push_back(i);
  }
}
vector<int> pos[maxn];
int has[maxn];
int f[maxn];
int pre[maxn];
int illpos;
set<int> fval;
inline long long solve() {
  long long ans = 0;
  for (int i = (1); i <= (n); ++i) dp[i] = n;
  for (int i = (0); i < (cnt); ++i) {
    if (prime[i] * prime[i] <= 1000000) {
      brute(prime[i]);
    } else
      break;
  }
  f[n + 1] = n;
  fval.insert(n + 1);
  illpos = n + 1;
  for (int i = (n); i >= (1); --i) {
    assert(vis[a[i]] == 0);
    if (a[i] == 1) {
      f[i] = f[i + 1];
    } else {
      if (b[i] == '/') {
        has[a[i]]--;
        pos[a[i]].push_back(-i);
        fval.erase(pre[a[i]]);
        pre[a[i]] = i;
        fval.insert(i);
        f[i] = i - 1;
      } else {
        if (!pos[a[i]].empty() && pos[a[i]].back() < 0) {
          pos[a[i]].pop_back();
          fval.erase(pre[a[i]]);
        }
        if (!pos[a[i]].empty() && pos[a[i]].back() < 0) {
          pre[a[i]] = -pos[a[i]].back();
          fval.insert(-pos[a[i]].back());
        }
        f[i] = (*fval.begin()) - 1;
      }
    }
  }
  cerr << f[1] << endl;
  for (int i = (1); i <= (n); ++i) {
    assert(min(f[i], dp[i]) + 1 >= i);
    ans += min(f[i], dp[i]) - i + 1;
  }
  return ans;
}
int main(void) {
  init();
  scanf("%d", &(n));
  for (int i = (1); i <= (n); ++i) scanf("%d", &(a[i]));
  scanf("%s", b + 1);
  printf("%lld\n", solve());
  return 0;
}

```