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#include <iostream>
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#include <vector>
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#include <queue>
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#include <cmath>
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using namespace std;
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const int N = (1 << 20) + 10;
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int n,m;
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int v[N],w[N];
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int res = -1;
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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
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struct good{
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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
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bool operator > (const good& W) const{
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return W.c + W.r > c + r;
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}
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}goods[N];
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int bfs(){
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goods[1] = {1,0,0,0};
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priority_queue<good,vector<good>,greater<good>> q;
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q.push(goods[1]);
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while(q.size()){
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auto t = q.top();
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// cout << t.idx << endl;
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q.pop();
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int idx = t.idx << 1;
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goods[idx] = {idx,goods[t.idx].c,s[n] - s[(int)log2(idx)],goods[t.idx].tv};
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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)]};
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if((int)log2(t.idx) == n) { // 假如已经是子节点，则更新答案
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res = max(res,t.c);
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continue;
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}
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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
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if(goods[idx + 1].tv <= m && goods[idx + 1].c + goods[idx + 1].r > res) q.push(goods[idx + 1]);
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}
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return res;
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}
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int main(){
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cin >> n >> m;
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for(int i = 1; i <= n; i ++) cin >> v[i] >> w[i],s[i] = s[i - 1] + w[i];
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//
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// for(int i = 2; i < 1 << n + 1; i ++){
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// 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)]};
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// }
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//
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// 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);
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cout << bfs() << endl;
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return 0;
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}
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