#include <bits/stdc++.h>
using namespace std;
const int N_MAX = 1000000;
const int A_MAX = 1000000;
int pdiv[A_MAX + 2];
int n;
int a[N_MAX + 2];
int sgn[N_MAX + 2];
vector<pair<int, short int>> muls[A_MAX + 2];
int SGT[N_MAX * 4 + 2];
void build(int node, int l, int r) {
  SGT[node] = n;
  if (l == r) return;
  int mid = (l + r) / 2;
  int lSon = node * 2, rSon = node * 2 + 1;
  build(lSon, l, mid);
  build(rSon, mid + 1, r);
}
void build() { build(1, 1, n); }
void update(int node, int l, int r, int ul, int ur, int uval) {
  if (ul <= l && r <= ur) {
    SGT[node] = min(SGT[node], uval);
    return;
  }
  int mid = (l + r) / 2;
  int lSon = node * 2, rSon = node * 2 + 1;
  if (ul <= mid) update(lSon, l, mid, ul, ur, uval);
  if (mid + 1 <= ur) update(rSon, mid + 1, r, ul, ur, uval);
}
void update(int ul, int ur, int uval) { update(1, 1, n, ul, ur, uval); }
int query(int node, int l, int r, int qpos) {
  if (l == r) return SGT[node];
  int mid = (l + r) / 2;
  int lSon = node * 2, rSon = node * 2 + 1;
  if (qpos <= mid)
    return min(SGT[node], query(lSon, l, mid, qpos));
  else
    return min(SGT[node], query(rSon, mid + 1, r, qpos));
}
int query(int qpos) { return query(1, 1, n, qpos); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (int p = 2; p <= A_MAX; p++)
    if (pdiv[p] == 0)
      for (int m = p; m <= A_MAX; m += p) pdiv[m] = p;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  string str;
  cin >> str;
  for (int i = 1; i <= n; i++) sgn[i] = (str[i - 1] == '*' ? +1 : -1);
  for (int i = 1; i <= n; i++) {
    int aux = a[i];
    while (aux > 1) {
      int p = pdiv[aux];
      int e = 0;
      do {
        e++;
        aux /= p;
      } while (aux % p == 0);
      muls[p].push_back(make_pair(i, e * sgn[i]));
    }
  }
  build();
  for (int p = 1; p <= A_MAX; p++)
    if (muls[p].empty() == false) {
      vector<int> pref;
      for (int k = 0; k < (int)muls[p].size(); k++) {
        pref.push_back((pref.empty() == true ? 0 : pref.back()));
        pref.back() += muls[p][k].second;
      }
      vector<int> st;
      for (int k = (int)muls[p].size() - 1; k >= 0; k--) {
        while (st.empty() == false && pref[k] <= pref[st.back()]) st.pop_back();
        int l = muls[p][k].first + 1;
        int r = (st.empty() == true ? n + 1 : muls[p][st.back()].first);
        update(l, r, r - 1);
        st.push_back(k);
      }
      while (st.empty() == false && 0 <= pref[st.back()]) st.pop_back();
      int l = 1;
      int r = (st.empty() == true ? n + 1 : muls[p][st.back()].first);
      update(l, r, r - 1);
    }
  long long answer = 0;
  for (int l = 1; l <= n; l++) answer += max(0, query(l) - l + 1);
  cout << answer << "\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;
const int N_MAX = 1000000;
const int A_MAX = 1000000;
int pdiv[A_MAX + 2];
int n;
int a[N_MAX + 2];
int sgn[N_MAX + 2];
vector<pair<int, short int>> muls[A_MAX + 2];
int SGT[N_MAX * 4 + 2];
void build(int node, int l, int r) {
  SGT[node] = n;
  if (l == r) return;
  int mid = (l + r) / 2;
  int lSon = node * 2, rSon = node * 2 + 1;
  build(lSon, l, mid);
  build(rSon, mid + 1, r);
}
void build() { build(1, 1, n); }
void update(int node, int l, int r, int ul, int ur, int uval) {
  if (ul <= l && r <= ur) {
    SGT[node] = min(SGT[node], uval);
    return;
  }
  int mid = (l + r) / 2;
  int lSon = node * 2, rSon = node * 2 + 1;
  if (ul <= mid) update(lSon, l, mid, ul, ur, uval);
  if (mid + 1 <= ur) update(rSon, mid + 1, r, ul, ur, uval);
}
void update(int ul, int ur, int uval) { update(1, 1, n, ul, ur, uval); }
int query(int node, int l, int r, int qpos) {
  if (l == r) return SGT[node];
  int mid = (l + r) / 2;
  int lSon = node * 2, rSon = node * 2 + 1;
  if (qpos <= mid)
    return min(SGT[node], query(lSon, l, mid, qpos));
  else
    return min(SGT[node], query(rSon, mid + 1, r, qpos));
}
int query(int qpos) { return query(1, 1, n, qpos); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (int p = 2; p <= A_MAX; p++)
    if (pdiv[p] == 0)
      for (int m = p; m <= A_MAX; m += p) pdiv[m] = p;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  string str;
  cin >> str;
  for (int i = 1; i <= n; i++) sgn[i] = (str[i - 1] == '*' ? +1 : -1);
  for (int i = 1; i <= n; i++) {
    int aux = a[i];
    while (aux > 1) {
      int p = pdiv[aux];
      int e = 0;
      do {
        e++;
        aux /= p;
      } while (aux % p == 0);
      muls[p].push_back(make_pair(i, e * sgn[i]));
    }
  }
  build();
  for (int p = 1; p <= A_MAX; p++)
    if (muls[p].empty() == false) {
      vector<int> pref;
      for (int k = 0; k < (int)muls[p].size(); k++) {
        pref.push_back((pref.empty() == true ? 0 : pref.back()));
        pref.back() += muls[p][k].second;
      }
      vector<int> st;
      for (int k = (int)muls[p].size() - 1; k >= 0; k--) {
        while (st.empty() == false && pref[k] <= pref[st.back()]) st.pop_back();
        int l = muls[p][k].first + 1;
        int r = (st.empty() == true ? n + 1 : muls[p][st.back()].first);
        update(l, r, r - 1);
        st.push_back(k);
      }
      while (st.empty() == false && 0 <= pref[st.back()]) st.pop_back();
      int l = 1;
      int r = (st.empty() == true ? n + 1 : muls[p][st.back()].first);
      update(l, r, r - 1);
    }
  long long answer = 0;
  for (int l = 1; l <= n; l++) answer += max(0, query(l) - l + 1);
  cout << answer << "\n";
  return 0;
}

```