#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;

const int N = (1 << 20) + 10;
int n,m;
int v[N],w[N];
int res = -1;
int s[N]; // s[i]存储的是从第1件物品到第i件物品的价值总和

struct good{
    int idx,c,r,tv; // idx表示选法下标，c表示该选法的当前总价值，r表示当前选法的剩余总价值，tv表示该选法的当前总体积
    bool operator > (const good& W) const{
        return W.c + W.r > c + r;
    }
}goods[N];

int bfs(){
    goods[1] = {1,0,0,0};
priority_queue<good,vector<good>,greater<good>> q;
    q.push(goods[1]);

    while(q.size()){
        auto t = q.top();
//        cout << t.idx << endl;
        q.pop();
        int idx = t.idx << 1;
        goods[idx] = {idx,goods[t.idx].c,s[n] - s[(int)log2(idx)],goods[t.idx].tv};
        goods[idx + 1] = {idx + 1,goods[t.idx].c + w[(int)log2(idx)],s[n] - s[(int)log2(idx)],goods[t.idx].tv + v[(int)log2(idx)]};
        if((int)log2(t.idx) == n) { // 假如已经是子节点，则更新答案
            res = max(res,t.c);
            continue;
        }

        if(goods[idx].tv <= m && goods[idx].c + goods[idx].r > res) q.push(goods[idx]); // 假如当前选法的总体积不超过背包容量，且当前价值+剩余价值 > 当前最优解，则装入背包
        if(goods[idx + 1].tv <= m && goods[idx + 1].c + goods[idx + 1].r > res) q.push(goods[idx + 1]);
    }

    return res;
}

int main(){
    cin >> n >> m;
    for(int i = 1; i <= n; i ++) cin >> v[i] >> w[i],s[i] = s[i - 1] + w[i];
//
//    for(int i = 2; i < 1 << n + 1; i ++){
//        goods[i] = {i,goods[i >> 1].c + (i&1)*w[(int)log2(i)],s[n] - s[(int)log2(i)],goods[i >> 1].tv + (i&1)*v[(int)log2(i)]};
//    }
//
//    for(int i = 1; i < 1 << n + 1; i ++) printf("i = %d,c[i] = %d,r[i] = %d,tv[i] = %d\n",goods[i].idx,goods[i].c,goods[i].r,goods[i].tv);
    cout << bfs() << endl;
    return 0;
}