Create 1922. Count Good Numbers

1922. Count Good Numbers

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+class Solution {
+public:
+    int M = 1e9+7;
+
+    long long binpow(long long a, long long b) {
+        a %= M;
+        long long res = 1;
+        while (b > 0) {
+            if (b & 1)
+                res = res * a % M;
+            a = a * a % M;
+            b >>= 1;
+        }
+        return res;
+    }
+
+    int countGoodNumbers(long long n) {
+        long long even_positions = (n + 1) / 2;
+        long long odd_positions = n / 2;
+        long long result = (binpow(5, even_positions) * binpow(4, odd_positions)) % M;
+        return result;
+    }
+};