// Original C++ solution (slower):
#ifdef _DEBUG

#include "MyLib.h"

#else

#define main_D main

#include "bits/stdc++.h" 

#include <regex>

#define _USE_MATH_DEFINES

#include <math.h>

#define FOR(i,s,e) for (int i = int(s); i < int(e); ++i)

#define REP(i,e) FOR(i,0,e)

#define INF (INT_MAX/2)

#define EPS (1.0e-8)

#define LINF (LLONG_MAX/2)

const int MGN = 8;

const int ARY_SZ_MAX = 10000000;

using namespace std;

using ll = long long; using ull = unsigned long long;

using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;

using vb = vector<bool>; using vvb = vector<vb>; using vvvb = vector<vvb>;

using vl = vector<ll>; using vvl = vector<vl>;

using vd = vector<double>; using vs = vector<string>;

using pii = pair<int, int>; using pll = pair<ll, ll>;

// functions



#endif





int main_D() {

    cin.tie(0);

    ios::sync_with_stdio(false);



    ll N, W; cin >> N >> W;

    vl w(N+1,0);

    vl v(N+1,0);

    FOR(i,1,N+1) cin >> w[i] >> v[i];



    ll WMIN = LINF;

    FOR(i,1,N+1) WMIN = min(WMIN, w[i]);    



    vi r(N+1,0);

    FOR(i, 1, N + 1) r[i] = w[i] - WMIN;



    const ll RMAX = 400;

    vector<vvl> dp(N+MGN, vvl(N+MGN, vl(RMAX+MGN, 0)));

    dp[0][0][0] = 0;



    FOR(i,1,N+1) {

        FOR(j,1,i+2) {

            for (ll k = 0; k <= RMAX; k++) {  

                ll next = (k >= r[i]) 

                    ? dp[i - 1][j - 1][k - r[i]] + v[i] : 0;

                dp[i][j][k] = max(next, dp[i - 1][j][k]);

            }

        }

    }



    ll ans = 0;

    for (ll j = 1; j <= N; j++) {

        if (WMIN * j <= W) {

            ans = max(ans, dp[N][j][min(RMAX, W - WMIN*j)]);

        }

    }



    cout << ans << endl;



    return 0;

}

// Optimized solution to the same problem (faster):
#ifdef _DEBUG

#include "MyLib.h"

#else

#define main_D main

#include "bits/stdc++.h" 

#include <regex>

#define _USE_MATH_DEFINES

#include <math.h>

#define FOR(i,s,e) for (int i = int(s); i < int(e); ++i)

#define REP(i,e) FOR(i,0,e)

#define INF (INT_MAX/2)

#define EPS (1.0e-8)

#define LINF (LLONG_MAX/2)

const int MGN = 8;

const int ARY_SZ_MAX = 10000000;

using namespace std;

using ll = long long; using ull = unsigned long long;

using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;

using vb = vector<bool>; using vvb = vector<vb>; using vvvb = vector<vvb>;

using vl = vector<ll>; using vvl = vector<vl>;

using vd = vector<double>; using vs = vector<string>;

using pii = pair<int, int>; using pll = pair<ll, ll>;

// functions



#endif





int main_D() {

    cin.tie(0);

    ios::sync_with_stdio(false);



    ll N, W; cin >> N >> W;

    vl w(N+1,0); vl v(N+1,0);

    FOR(i,1,N+1) cin >> w[i] >> v[i];



    ll WMIN = LINF;

    FOR(i,1,N+1) WMIN = min(WMIN, w[i]);    



    vi r(N+1,0);

    FOR(i,1,N + 1) r[i] = w[i] - WMIN;



    vi rs(N+1, 0);

    FOR(i,1,N + 1) rs[i] = rs[i - 1] + r[i];



    const ll RMAX = 300;

    vvl dp(N+MGN, vl(RMAX+MGN, 0));

    // dp[j][k] := max value when taking j items from {s1,s2,...,si} with r-weight le k 



    FOR(i,0,N+1) {

        for(int j=i; j>=0; j--) {

            for (int k = rs[i]; k >= 0; k--) {

                assert(k + r[i] < dp[0].size());

                dp[j+1][k+r[i]] = max(dp[j+1][k+r[i]], dp[j][k]+v[i]);

            }

        }

    }



    ll ans = 0;

    FOR(i,0,N+1) {

        FOR(j,0,rs[N]+1) {

            ll w = WMIN * i + j;

            if (w <= W){

                assert(j < dp[0].size());

                ans = max(ans, dp[i][j]);

            }

        }

    }



    cout << ans << endl;



    return 0;

}

// Fastest known solution for this problem: