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Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n class Solution {\nprivate:\nlong long pow(long x,long y,int mod)\n{\n    if(y==0)\n    {\n        return 1;\n    }\n    long long ans=pow(x,y/2,mod);\n    if(y%2==0)\n    {\n        return (ans*ans)%mod;\n    }\n    return ((((x*ans)%mod)*ans)%mod);\n}\npublic:\n    int stringCount(int n) {\n        long count = 0;\n        int mod = 1e9 + 7;\n        //using inclusin exclusion principle\n        //case c1 represents string with zero l\n        //case c2 represents string with zero e\n        //case c3 represents strings with 0 t\n        //case c4 represents case with 1 e\n        //i want union of (c1,c2,c3,c4)\n        //calculation it using inclusion exclusion principle \n        //union of(c1,c2,c3,c4)->tells about the bad strings=   c1+c2+c3+c4 -(I(c1,c2)+I(c2,c3)+I(c3,c4)+I(c4,c1))+(I(C1,c2,c3)+I(c2,c3,c4)+I(c3,c4,c1)+I(c4,c1,c2))-(I(c1,c2,c3,c4));\n        //I(c1,c2) represents intersection of c1 and c2\n        //pow(26,n,mod)->tells about the number of strings \n        return ((pow(26, n, mod)- (n + 75) * pow(25, n - 1, mod)+ (2 * n + 72) * pow(24, n - 1, mod)- (n + 23) * pow(23, n - 1, mod)) % mod + mod) % mod;\n    }\n};\n```
['C++']