1 #include <stdio.h>
2 #include <algorithm>
3 using namespace std;
4
5 typedef long long ll;
6 const int mod = 1e9 + 7;
7 int dp[1010][10000];
8 // dp[i][j] : i 个数,组成总和为j 的数量
9
10 int main()
11 {
12 int n;
13 scanf("%d", &n);
14 dp[0][0] = 1;
15 for (int i = 0; i <= 9; i++)
16 dp[1][i] = 1;
17 for (int i = 2; i <= n; i++)
18 {
19 for (int j = 0; j <= 9 * i; j++)
20 {
21 int sum = 0;
22 for (int k = 0; k <= 9; k++)
23 {
24 if (j >= k)
25 sum = (sum + dp[i - 1][j - k]) % mod;
26 else
27 break;
28 }
29 dp[i][j] = sum;
30 }
31 }
32 ll ans = 0;
33 for (int i = 0; i <= 9 * n; i++)
34 ans = (ans + (ll)dp[n][i] * (dp[n][i] - dp[n - 1][i])) % mod;
35 printf("%lld\n", ans);
36 return 0;
37 }