#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll NEG = (ll)-4e18;
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, T;
    if (!(cin >> n >> T)) return 0;
    vector<int> t(n), q(n);
    for (int i = 0; i < n; ++i) cin >> t[i] >> q[i];
 
    // group tasks by cap = T - t[i]
    vector<vector<int>> byCap(T+1);
    for (int i = 0; i < n; ++i) {
        int cap = T - t[i];
        if (cap >= 0 && cap <= T) byCap[cap].push_back(q[i]);
    }
    // sort each level decreasing and build prefix sums
    vector<vector<ll>> pref(T+1);
    for (int h = 0; h <= T; ++h) {
        auto &v = byCap[h];
        sort(v.begin(), v.end(), greater<int>());
        pref[h].assign(v.size()+1, 0);
        for (size_t k = 1; k <= v.size(); ++k) pref[h][k] = pref[h][k-1] + v[k-1];
    }
 
    int MAXC = n; // số nút ở 1 mức không vượt quá n
    // D[h][c] : giá trị tối đa đã thu được từ mức > h khi hiện có c nút ở mức h
    vector<vector<ll>> D(T+1, vector<ll>(MAXC+1, NEG));
 
    // Khởi: ở mức sâu nhất (h = T) "không có cấp sâu hơn", nên đóng góp dưới cùng = 0
    // Ta cho D[T][c] = 0 cho mọi c (tức chưa thu gì từ dưới)
    for (int c = 0; c <= MAXC; ++c) D[T][c] = 0;
 
    // sweep từ h = T xuống h = 0
    ll answer = 0; // có thể là 0 (không chọn task nào)
    for (int h = T; h >= 0; --h) {
        // nếu h == 0 thì khi cập nhật lên h-1 thực tế là kết thúc -> ta thu kết quả vào answer
        for (int c = 0; c <= MAXC; ++c) {
            if (D[h][c] == NEG) continue;
            int maxk = min((int)byCap[h].size(), c);
            for (int k = 0; k <= maxk; ++k) {
                ll val = D[h][c] + pref[h][k];
                int newC = (c + k + 1) / 2; // ceil((c+k)/2)
                if (h == 0) {
                    // ở mức 0: muốn có đúng 1 nút (root). newC phải là 1
                    if (newC == 1) answer = max(answer, val);
                } else {
                    // cập nhật cho mức trên (h-1)
                    D[h-1][newC] = max(D[h-1][newC], val);
                }
            }
        }
    }
 
    cout << answer << '\n';
    return 0;
}
 

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