#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chkMax(T &x, T y) {
  if (y > x) x = y;
}
template <typename T>
void chkMin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
const int N = 1e6 + 5;
int n, a[N], p0[N];
void inline prework() {
  int n = 1e6;
  for (int i = 2; i <= n; i++) {
    for (int j = i; j <= n; j += i)
      if (!p0[j]) p0[j] = i;
  }
}
vector<pair<int, int> > e[N];
char g[N];
int s[N], top, sum[N];
vector<pair<int, int> > q[N];
vector<pair<int, int> > G;
void inline work(vector<pair<int, int> > &t) {
  int m = t.size();
  if (!m) return;
  G.clear();
  G.push_back(make_pair(0, 0));
  for (int i = 0; i < m; i++) G.push_back(t[i]);
  for (int i = 1; i <= m; i++) {
    sum[i] = G[i].first + sum[i - 1];
  }
  top = 0;
  for (int i = m; i >= 0; i--) {
    while (top && sum[s[top]] >= sum[i]) top--;
    if (s[top]) {
      q[G[i].second + 1].push_back(make_pair(G[s[top]].second, 1));
      if (i != m)
        q[G[i + 1].second + 1].push_back(make_pair(G[s[top]].second, -1));
    }
    s[++top] = i;
  }
}
multiset<int> S;
int main() {
  prework();
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  scanf("%s", g + 1);
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    int w = g[i] == '*' ? 1 : -1;
    while (x > 1) {
      int v = p0[x], c = 0;
      while (x % v == 0) x /= v, c++;
      e[v].push_back(make_pair(w * c, i));
    }
  }
  long long ans = 0;
  for (int i = 1; i <= 1000000; i++) work(e[i]);
  for (int i = 1; i <= n; i++) {
    for (pair<int, int> u : q[i]) {
      if (u.second == 1)
        S.insert(u.first);
      else
        S.erase(S.find(u.first));
    }
    int lim = n;
    if (S.size()) lim = *S.begin() - 1;
    if (i <= lim) ans += lim - i + 1;
  }
  printf("%lld\n", ans);
  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;
template <typename T>
void chkMax(T &x, T y) {
  if (y > x) x = y;
}
template <typename T>
void chkMin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
const int N = 1e6 + 5;
int n, a[N], p0[N];
void inline prework() {
  int n = 1e6;
  for (int i = 2; i <= n; i++) {
    for (int j = i; j <= n; j += i)
      if (!p0[j]) p0[j] = i;
  }
}
vector<pair<int, int> > e[N];
char g[N];
int s[N], top, sum[N];
vector<pair<int, int> > q[N];
vector<pair<int, int> > G;
void inline work(vector<pair<int, int> > &t) {
  int m = t.size();
  if (!m) return;
  G.clear();
  G.push_back(make_pair(0, 0));
  for (int i = 0; i < m; i++) G.push_back(t[i]);
  for (int i = 1; i <= m; i++) {
    sum[i] = G[i].first + sum[i - 1];
  }
  top = 0;
  for (int i = m; i >= 0; i--) {
    while (top && sum[s[top]] >= sum[i]) top--;
    if (s[top]) {
      q[G[i].second + 1].push_back(make_pair(G[s[top]].second, 1));
      if (i != m)
        q[G[i + 1].second + 1].push_back(make_pair(G[s[top]].second, -1));
    }
    s[++top] = i;
  }
}
multiset<int> S;
int main() {
  prework();
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  scanf("%s", g + 1);
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    int w = g[i] == '*' ? 1 : -1;
    while (x > 1) {
      int v = p0[x], c = 0;
      while (x % v == 0) x /= v, c++;
      e[v].push_back(make_pair(w * c, i));
    }
  }
  long long ans = 0;
  for (int i = 1; i <= 1000000; i++) work(e[i]);
  for (int i = 1; i <= n; i++) {
    for (pair<int, int> u : q[i]) {
      if (u.second == 1)
        S.insert(u.first);
      else
        S.erase(S.find(u.first));
    }
    int lim = n;
    if (S.size()) lim = *S.begin() - 1;
    if (i <= lim) ans += lim - i + 1;
  }
  printf("%lld\n", ans);
  return 0;
}

```