## solution

0がなければ結果の整数は111...1222...2333...3.........999...9という形。 ある場合は111000...0111...1222...2.........222000...0222...2333...3.........のように必要最小限だけ前に出す。 これらをいい感じに計算すればよい。 repunit $111 \dots 1$や$10^k$は線形の漸化式を持つので桁数の対数で求められる。階乗する部分が一番重くて$O(N + \sum b_i)$。

## implementation

#include <array>
#include <cassert>
#include <cstdio>
#include <tuple>
#include <vector>
#define repeat(i, n) for (int i = 0; (i) < int(n); ++(i))
#define repeat_from(i, m, n) for (int i = (m); (i) < int(n); ++(i))
using ll = long long;
using namespace std;

template <int mod>
int fact(int n) {
static vector<int> memo(1, 1);
if (memo.size() <= n) {
int l = memo.size();
memo.resize(n+1);
repeat_from (i, l, n+1) memo[i] = memo[i-1] *(ll) i % mod;
}
return memo[n];
}
ll powmod(ll x, ll y, ll p) { // O(log y)
assert (0 <= x and x < p);
assert (0 <= y);
ll z = 1;
for (ll i = 1; i <= y; i <<= 1) {
if (y & i) z = z * x % p;
x = x * x % p;
}
return z;
}
int repunit(ll n, int mod) { // O(log n)
ll y = 0;
ll x = 1;
for (ll i = 1; i <= n; i <<= 1) {
if (n & i) y = (y * powmod(10, i, mod) % mod + x) % mod;
x = (x * powmod(10, i, mod) % mod + x) % mod;
}
return y;
}

constexpr int mod = 1e9+7;
pair<int, int> solve(int n, vector<int> const & a, vector<int> const & b) {
array<int, 10> cnt = {};
array<ll, 10> acc = {};
repeat (i, n) {
cnt[a[i]] += 1;
acc[a[i]] += b[i];
}
int x = 0;
int y = 1;
if (cnt[0]) {
if (n == 1) return make_pair(0, 1);
int d = 1;
while (not cnt[d]) ++ d;
assert (d <= 9);
int min_length = mod;
int min_length_count = 0;
repeat (i, n) {
if (a[i] == d and b[i] <= min_length) {
if (b[i] < min_length) {
min_length = b[i];
min_length_count = 0;
}
min_length_count += 1;
}
}
x = (x *(ll) powmod(10, min_length, mod) + repunit(min_length, mod) *(ll) d % mod) % mod;
y = y *(ll) min_length_count % mod;
cnt[d] -= 1;
acc[d] -= min_length;
}
repeat (d, 10) {
x = (x *(ll) powmod(10, acc[d], mod) + repunit(acc[d], mod) *(ll) d % mod) % mod;
y = y *(ll) fact<mod>(cnt[d]) % mod;
}
return make_pair(x, y);
}

int main() {
int n; scanf("%d", &n);
vector<int> a(n), b(n); repeat (i, n) scanf("%d%d", &a[i], &b[i]);
int x, y; tie(x, y) = solve(n, a, b);
printf("%d\n", x);
printf("%d\n", y);
return 0;
}