#include <bits/stdc++.h>
using namespace std;

int n;
vector<string> a;   // giu nguyen dang xau de lay chu so dau va chu so cuoi (a_i <= 10^18)

// Sub 1: vet can O(n^2) - xet tung cap (i, j) voi i < j, kiem tra ucln(chu so dau cua a_i, chu so cuoi cua a_j) = 1.
// Dung khi n <= 1000.
void sub1() {
    long long ans = 0;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++) {
            int first = a[i][0] - '0';
            int last = a[j].back() - '0';
            if (__gcd(first, last) == 1) ans++;
        }
    cout << ans << "\n";
}

// Sub 2: duyet j tu trai sang phai, cntFirst[d] = so i < j co chu so dau la d.
// Voi moi j, cong cntFirst[d] voi moi d nguyen to cung nhau voi chu so cuoi cua a_j (chi 9 gia tri d). O(9n).
void sub2() {
    long long cntFirst[10] = {0};
    long long ans = 0;
    for (int j = 0; j < n; j++) {
        int last = a[j].back() - '0';
        for (int d = 1; d <= 9; d++)
            if (__gcd(d, last) == 1) ans += cntFirst[d];
        cntFirst[a[j][0] - '0']++;      // them a_j sau khi xet, de chi tinh cap i < j
    }
    cout << ans << "\n";
}

int main() {
    freopen("CAU2.INP", "r", stdin);
    freopen("CAU2.OUT", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n;
    a.resize(n);
    for (auto& x : a) cin >> x;
    if (n <= 1000) sub1();
    else sub2();
    return 0;
}