// 数位DP（非递归）

// https://www.luogu.com.cn/problem/P13085

#include <bits/stdc++.h>

using namespace std;

using ll = long long;

/**

* @brief 非递归数位DP

* @param num 上界

* @return 满足题意的数的个数

*/

ll f(ll num) {

vector<int> arr(19);

for (int i = 0; i < 19; i++) {

arr[i] = num % 10;

num /= 10;

}

// dp[位置][末尾的数][是否遇到非零][是否限制]

ll dp[20][10][2][2]{};

dp[19][0][0][1] = 1;

for (int p = 18; p >= 0; p--)

for (int x = 0; x < 10; x++)

for (int s = 0; s < 2; s++)

for (int l = 0; l < 2; l++) {

int up = l ? arr[p] : 9;

for (int y = 0; y <= up; y++) {

if (!s || abs(y - x) >= 2)

dp[p][y][s || y][l && y == up] += dp[p + 1][x][s][l];

}

}

ll ans = 0;

for (int x = 0; x < 10; x++)

for (int s = 0; s < 2; s++)

for (int l = 0; l < 2; l++) ans += dp[0][x][s][l];

return ans;

}

int main() {

ll a, b;

cin >> a >> b;

ll ans1 = f(a - 1);

ll ans2 = f(b);

cout << ans2 - ans1 << "\n";

return 0;

}