#include <bits/stdc++.h>
using namespace std;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
int N;
int A[1010101];
vector<int> V[1010101];
string S;
const int prime_max = 1000;
vector<int> prime;
int NP, divp[prime_max];
void cprime() {
  if (NP) return;
  for (int i = 2; i < prime_max; i++)
    if (divp[i] == 0) {
      prime.push_back(i);
      NP++;
      for (signed long long j = 1LL * i * i; j >= i && j < prime_max; j += i)
        if (divp[j] == 0) divp[j] = i;
    }
}
int R[1010101];
template <class V, int NV>
class SegTree_2 {
 public:
  V nolink;
  vector<V> val;
  SegTree_2() {
    val.resize(NV * 2);
    clear();
    nolink = 1 << 30;
  };
  void clear() {
    for (int i = 0; i < NV * 2; i++) val[i] = 1 << 20;
  }
  V getval(int k) {
    int e = k + NV;
    V ret = nolink;
    while (e >= 1) ret = min(ret, val[e]), e /= 2;
    return ret;
  }
  void update(int x, int y, V v, int l = 0, int r = NV, int k = 1) {
    if (l >= r) return;
    if (x <= l && r <= y)
      val[k] = min(val[k], v);
    else if (l < y && x < r) {
      update(x, y, v, l, (l + r) / 2, k * 2);
      update(x, y, v, (l + r) / 2, r, k * 2 + 1);
    }
  }
};
SegTree_2<int, 1 << 20> st;
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cprime();
  cin >> N;
  for (i = 0; i < (N); i++) {
    cin >> x;
    for (auto &p : prime)
      while (x % p == 0) {
        V[p].push_back(i);
        x /= p;
      }
    if (x > 1) V[x].push_back(i);
  }
  st.update(0, N, N - 1);
  cin >> S;
  for (i = 0; i < (1010000); i++)
    if (V[i].size()) {
      vector<int> A;
      for (auto &v : V[i]) {
        if (S[v] == '*') {
          A.push_back(v);
        } else {
          if (A.size()) {
            x = A.back();
            A.pop_back();
          } else {
            x = -1;
          }
          st.update(x + 1, v + 1, v - 1);
        }
      }
    }
  signed long long ret = 0;
  for (i = 0; i < (N); i++) {
    x = st.getval(i);
    ret += (x + 1) - i;
  }
  cout << ret << endl;
}
int main(int argc, char **argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Please create a solution in CPP 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;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
int N;
int A[1010101];
vector<int> V[1010101];
string S;
const int prime_max = 1000;
vector<int> prime;
int NP, divp[prime_max];
void cprime() {
  if (NP) return;
  for (int i = 2; i < prime_max; i++)
    if (divp[i] == 0) {
      prime.push_back(i);
      NP++;
      for (signed long long j = 1LL * i * i; j >= i && j < prime_max; j += i)
        if (divp[j] == 0) divp[j] = i;
    }
}
int R[1010101];
template <class V, int NV>
class SegTree_2 {
 public:
  V nolink;
  vector<V> val;
  SegTree_2() {
    val.resize(NV * 2);
    clear();
    nolink = 1 << 30;
  };
  void clear() {
    for (int i = 0; i < NV * 2; i++) val[i] = 1 << 20;
  }
  V getval(int k) {
    int e = k + NV;
    V ret = nolink;
    while (e >= 1) ret = min(ret, val[e]), e /= 2;
    return ret;
  }
  void update(int x, int y, V v, int l = 0, int r = NV, int k = 1) {
    if (l >= r) return;
    if (x <= l && r <= y)
      val[k] = min(val[k], v);
    else if (l < y && x < r) {
      update(x, y, v, l, (l + r) / 2, k * 2);
      update(x, y, v, (l + r) / 2, r, k * 2 + 1);
    }
  }
};
SegTree_2<int, 1 << 20> st;
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cprime();
  cin >> N;
  for (i = 0; i < (N); i++) {
    cin >> x;
    for (auto &p : prime)
      while (x % p == 0) {
        V[p].push_back(i);
        x /= p;
      }
    if (x > 1) V[x].push_back(i);
  }
  st.update(0, N, N - 1);
  cin >> S;
  for (i = 0; i < (1010000); i++)
    if (V[i].size()) {
      vector<int> A;
      for (auto &v : V[i]) {
        if (S[v] == '*') {
          A.push_back(v);
        } else {
          if (A.size()) {
            x = A.back();
            A.pop_back();
          } else {
            x = -1;
          }
          st.update(x + 1, v + 1, v - 1);
        }
      }
    }
  signed long long ret = 0;
  for (i = 0; i < (N); i++) {
    x = st.getval(i);
    ret += (x + 1) - i;
  }
  cout << ret << endl;
}
int main(int argc, char **argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}

```