Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example, if x?=?4, and y?=?7, then numbers 47, 744, 4 are lucky.

Let’s call a positive integer a undoubtedly lucky, if there are such digits x and y (0?≤?x,?y?≤?9), that the decimal representation of number a (without leading zeroes) contains only digits x and y.

Polycarpus has integer n. He wants to know how many positive integers that do not exceed n, are undoubtedly lucky. Help him, count this number.

Input
The first line contains a single integer n (1?≤?n?≤?109) — Polycarpus’s number.

Output
Print a single integer that says, how many positive integers that do not exceed n are undoubtedly lucky.

Examples
Input
10
Output
10
Input
123
Output
113
Note
In the first test sample all numbers that do not exceed 10 are undoubtedly lucky.

In the second sample numbers 102, 103, 104, 105, 106, 107, 108, 109, 120, 123 are not undoubtedly lucky.

給一個數n，求小于n但是數位相差不能超過二的數字。既然不能超過2，就兩邊循環。再加上dfs就可以跑出來了。
代碼如下：

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<set> #define ll long long using namespace std;ll n; set<ll> s;//定義一個集合，就可以快速去重void dfs(int x,int y,ll ans) {s.insert(ans);ll tx=10*ans+x;ll ty=10*ans+y;if(tx&&tx<=n)//0除外dfs(x,y,tx);if(ty&&ty<=n)dfs(x,y,ty); } int main() {while(scanf("%I64d",&n)!=EOF){s.clear();for(int i=0;i<=9;i++)for(int j=0;j<=9;j++){dfs(i,j,0);}printf("%lld\n",s.size()-1);//n除外} }

努力加油a啊，(^o)/~

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