#include <bits/stdc++.h>
using std::pair;
using std::vector;
const int MAX_N = 1000005, MAX_V = 1000005, MAX_PN = 78499;
int v[MAX_V], p[MAX_PN], pn = 0;
int a[MAX_N];
char opt[MAX_N];
int req[MAX_N];
int n;
vector<int> vec[MAX_V];
pair<int, int> stk[MAX_N];
inline void chkmn(int &a, int b) {
  if (a > b) a = b;
}
void eulerSieve(int n = 1000000) {
  for (int i = 2; i <= n; ++i) {
    if (!v[i]) v[i] = p[++pn] = i;
    for (int j = 1; 1ll * i * p[j] <= n; ++j) {
      v[i * p[j]] = p[j];
      if (i % p[j] == 0) break;
    }
  }
}
void ins(int pos) {
  req[pos] = pos;
  for (int x = a[pos]; x > 1; x /= v[x]) vec[v[x]].push_back(pos);
}
void del(int pos) {
  req[pos] = pos;
  for (int x = a[pos]; x > 1; x /= v[x]) {
    if (vec[v[x]].empty()) {
      req[pos] = 0;
      return;
    }
    chkmn(req[pos], vec[v[x]].back());
    vec[v[x]].pop_back();
  }
}
int main() {
  eulerSieve();
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= n; ++i) scanf(" %c", opt + i);
  for (int i = 1; i <= n; ++i)
    if (opt[i] == '*')
      ins(i);
    else
      del(i);
  long long ans = 0;
  for (int i = n, top = 0; i; --i) {
    int cnt = 1;
    while (top and req[i] <= stk[top].first) cnt += stk[top--].second;
    stk[++top] = pair<int, int>(req[i], cnt);
    if (req[i] == i) ans += cnt;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 std::pair;
using std::vector;
const int MAX_N = 1000005, MAX_V = 1000005, MAX_PN = 78499;
int v[MAX_V], p[MAX_PN], pn = 0;
int a[MAX_N];
char opt[MAX_N];
int req[MAX_N];
int n;
vector<int> vec[MAX_V];
pair<int, int> stk[MAX_N];
inline void chkmn(int &a, int b) {
  if (a > b) a = b;
}
void eulerSieve(int n = 1000000) {
  for (int i = 2; i <= n; ++i) {
    if (!v[i]) v[i] = p[++pn] = i;
    for (int j = 1; 1ll * i * p[j] <= n; ++j) {
      v[i * p[j]] = p[j];
      if (i % p[j] == 0) break;
    }
  }
}
void ins(int pos) {
  req[pos] = pos;
  for (int x = a[pos]; x > 1; x /= v[x]) vec[v[x]].push_back(pos);
}
void del(int pos) {
  req[pos] = pos;
  for (int x = a[pos]; x > 1; x /= v[x]) {
    if (vec[v[x]].empty()) {
      req[pos] = 0;
      return;
    }
    chkmn(req[pos], vec[v[x]].back());
    vec[v[x]].pop_back();
  }
}
int main() {
  eulerSieve();
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= n; ++i) scanf(" %c", opt + i);
  for (int i = 1; i <= n; ++i)
    if (opt[i] == '*')
      ins(i);
    else
      del(i);
  long long ans = 0;
  for (int i = n, top = 0; i; --i) {
    int cnt = 1;
    while (top and req[i] <= stk[top].first) cnt += stk[top--].second;
    stk[++top] = pair<int, int>(req[i], cnt);
    if (req[i] == i) ans += cnt;
  }
  printf("%lld\n", ans);
  return 0;
}

```