#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <map>
#include <set>
#include <bitset>
#include <sstream>
#include <cstdlib>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <climits>
#include <ctype.h>
using namespace std;

#define PI acos(-1.0)
#define EPS 1e-8
#define MOD 1000000007
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define fi first
#define se second

typedef long double ld;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;

struct Matrix {
    vvi data;
    int r, c;
    Matrix(int row, int col, bool identity = false) : r(row), c(col) {
        data.assign(row, vector<int>(col, 0));
        if (identity) {
            for (int i = 0; i < r; ++i) {
                data[i][i] = 1;
            }
        }
    }
    Matrix operator * (Matrix& other) {
        int m = r, n = c, p = other.c;
        Matrix res(m, p);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < p; ++j) {
                for (int k = 0; k < n; ++k) {
                    res.data[i][j] += data[i][k] * other.data[k][j];
                }
            }
        }
        return res;
    }
};

// 扩展欧几里得算法 res[1] * a + res[2] * b = res[0] = gcd(a, b)
vector<int> extendGcd(int a, int b) {
    if (b == 0) {
        return {a, 1, 0};
    } else {
        vector<int> tmp = extendGcd(b, a % b);
        return {tmp[0], tmp[2], tmp[1] - (a / b) * tmp[2]};
    }
}

// 矩阵快速幂
Matrix matrix_power(Matrix base, ll exp) {
    int n = base.r;
    Matrix res(n, n, true);
    while (exp) {
        if (exp & 1) {
            res = res * base;
        }
        base = base * base;
        exp >>= 1;
    }
    return res;
}

// 带模快速幂
template <typename T>
T power_mod(T base, int exp, int mod = MOD) {
    T res = 1;
    while (exp) {
        if (exp & 1) res = res * base % mod;
        base = base * base % mod;
        exp >>= 1;
    }
    return res;
}

// 快速幂
template <typename T>
T power(T base, int exp) {
    T res = 1;
    while (exp) {
        if (exp & 1) res *= base;
        base *= base;
        exp >>= 1;
    }
    return res;
}

// 求逆元
ll inv(ll a) {
    return power_mod(a, MOD - 2);
}

/******************************** template ********************************/
#define N 2010
int n;
int z, w;
int a[N];
unordered_map<int, int> lastPos;
map<pii, int> maxDp, minDp;

int solveMin(int ind, int other);

int solveMax(int ind, int other) {
    if (maxDp.find({ind, other}) != maxDp.end()) {
        return maxDp[{ind, other}];
    }
    // draw all remaining cards
    int res = abs(other - a[n - 1]);
    for (int i = ind; i < n - 1; ++i) {
        res = max(res, solveMin(i + 1, a[i]));
    }
    return maxDp[{ind, other}] = res;
}

int solveMin(int ind, int other) {
    if (minDp.find({ind, other}) != minDp.end()) {
        return minDp[{ind, other}];
    }
    // draw all remaining cards
    int res = abs(other - a[n - 1]);
    for (int i = ind; i < n - 1; ++i) {
        res = min(res, solveMax(i + 1, a[i]));
    }
    return minDp[{ind, other}] = res;
}

int main() {
    cin >> n >> z >> w;
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        lastPos[a[i]] = i;
    }
    cout << solveMax(0, w) << endl;

    return 0;
}

Measured memory_bytes: 768