Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

2 60
10 10
100 100

Sample Output 1

110

By ordering the first and second dishes in this order, Takahashi's happiness will be 110.

Note that, if we manage to order a dish in time, we can spend any amount of time to eat it.

Sample Input 2

3 60
10 10
10 20
10 30

Sample Output 2

60

Takahashi can eat all the dishes within 60 minutes.

Sample Input 3

3 60
30 10
30 20
30 30

Sample Output 3

50

By ordering the second and third dishes in this order, Takahashi's happiness will be 50.

We cannot order three dishes, in whatever order we place them.

Sample Input 4

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

Sample Output 4

145

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// CODE by Kazhybay Askar

//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>

using namespace std;
//using namespace __gnu_pbds;

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC    optimize("Ofast")
//#pragma GCC    target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define iter   set<int>::iterator
#define f      first
#define s      second
#define pb     push_back
#define pf     push_front
#define ppb    pop_back
#define ppf    pop_front
#define mp     make_pair
#define cont   continue
#define Speed  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define sz     size()
#define all(x) x.begin(),x.end()
#define cont   continue
#define r0     return 0
#define endl   '\n'

typedef long long ll;
typedef long double   ld;
typedef pair<ll,ll>   pll;
typedef pair<int,int> pii;
typedef pair<char,char> pcc;

const int MAXN  = 5e5  + 7;
const int MOD   = 1e9  + 7;
const int INF   = 1e9  + 7;
const ll  MINF  = 1e18 + 7;
const int EPS   = 1e-6;
const int dx[]  = {0, 0, 1, -1, -1, -1, 1, 1};
const int dy[]  = {1, -1, 0, 0, -1, 1, -1, 1};
const double pi = acos(-1.0);

void freopen(){
    freopen("A.in","r",stdin);
    freopen("A.out","w",stdout);
}

ll n,mx,s,dp[MAXN];


void solve(){
    cin>>n>>s;
    for(int i=1;i<=n;i++){
        ll a,b;
        cin>>a>>b;
        for(int j=6001;j>=0;j--){
            if(j-a<s && 0<=j-a){
                dp[j]=max(dp[j-a]+b,dp[j]);
                //cout<<111;
            }
            else dp[j]=max(dp[j+1],dp[j]);
            mx=max(mx,dp[j]);
        }
    }
    cout<<mx;
}

int main(){
    Speed;
    ll T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    r0;
}