Problem: Takahashi is at an all-you-can-eat restaurant.

The restaurant offers N kinds of dishes. It takes A_i minutes to eat the i-th dish, whose deliciousness is B_i.

The restaurant has the following rules:

* You can only order one dish at a time. The dish ordered will be immediately served and ready to eat.
* You cannot order the same kind of dish more than once.
* Until you finish eating the dish already served, you cannot order a new dish.
* After T-0.5 minutes from the first order, you can no longer place a new order, but you can continue eating the dish already served.



Let Takahashi's happiness be the sum of the deliciousness of the dishes he eats in this restaurant.

What is the maximum possible happiness achieved by making optimal choices?

Constraints

* 2 \leq N \leq 3000
* 1 \leq T \leq 3000
* 1 \leq A_i \leq 3000
* 1 \leq B_i \leq 3000
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N T
A_1 B_1
:
A_N B_N


Output

Print the maximum possible happiness Takahashi can achieve.

Examples

Input

2 60
10 10
100 100


Output

110


Input

3 60
10 10
10 20
10 30


Output

60


Input

3 60
30 10
30 20
30 30


Output

50


Input

10 100
15 23
20 18
13 17
24 12
18 29
19 27
23 21
18 20
27 15
22 25


Output

145

Solution (in JAVA):

Solution:
import java.io.PrintWriter;
import java.util.*;

public class Main {
    PrintWriter out = new PrintWriter(System.out);
    final long mod = 1000000000+7;
    public static void main(String[] args) {
        Main main = new Main();
        System.out.println(main.solve());
    }
    // segment tree to record minimum value
    // Each point has a infinite distance from 0 (except itself)
    long solve() {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), T = sc.nextInt();
        int[][] weight_value = new int[n][2];
        for(int i=0;i<n;i++){
            int w = sc.nextInt(), v = sc.nextInt();
            weight_value[i][0] = w; weight_value[i][1] = v;
        }
        // put the one with smaller weight at front
        Arrays.sort(weight_value,(o1,o2)->(o1[0]==o2[0]?o1[1]-o2[1]:o1[0]-o2[0]));
        // note that traditional back-pack ends at sum(weight)=T-1
        int[][] dp = new int[n+1][T+1];
        for(int i=1;i<=n;i++){
            int w = weight_value[i-1][0], v = weight_value[i-1][1];
            int maxi = 0;
            for(int j=0;j<T;j++){
                dp[i][j] = dp[i-1][j];
                if(j>=w){
                    dp[i][j] = Math.max(dp[i][j], dp[i-1][j-w]+v);
                }
                maxi = Math.max(maxi, dp[i-1][j]);
            }
            // finally the tricky "tail"
            dp[i][T] = Math.max(dp[i-1][T], maxi+v);
        }
        long ans = 0;
        for(int j=0;j<=T;j++) ans = Math.max(ans, dp[n][j]);
        return ans;
    }
}