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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

// AtCoder.cpp : このファイルには 'main' 関数が含まれています。プログラム実行の開始と終了がそこで行われます。
//

// AtCoder.cpp : このファイルには 'main' 関数が含まれています。プログラム実行の開始と終了がそこで行われます。
//

#ifndef ONLINE_JUDGE

#endif
#define _USE_MATH_DEFINES
#include<math.h>

//#include<cmath>

#include<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<bitset>
#include<stack>
#include<memory>
#pragma GCC optimize("Ofast")


//#include "Ants.h"
using namespace std;
typedef long long ll;
#define rad_to_deg(rad) (((rad)/2/M_PI)*360)
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define rep(i,n) for(int i=0;i<n;i++)
#define show(s) cout<<s<<endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
#define LINF (1LL << 50)
#define MOD (1e9+7)
#define rep1(i,n) for(int i =1;i<=n;i++)
#define rrep(i,n) for(int i=n-1;i>=0;--i)

typedef pair < ll, ll> P;
ll gcd(ll a, ll b) {//最大公約数を求める///最大公倍数a*b/gcd(a,b)
	if (b == 0) return a;
	return gcd(b, a % b);
}
ll lcd(ll a, ll b) {//最小公倍数を求める　b=0のときaを返す
	if (b == 0) return a;
	return (a / gcd(a, b)) * b;
}
vector<int> divnum(ll num) {
	int dig;
	vector<int>p;
	while (num) {
		dig = num % 10;
		p.push_back(dig);
		num /= 10;
	}
	return p;
}
int digiter(ll num) {
	int dig;
	vector<int>p;
	while (num) {
		dig = num % 10;
		p.push_back(dig);
		num /= 10;
	}
	return p.size();
}
vector<int> convertstring(string s) {
	vector<int> d;
	ll n = s.size();
	rep(i, n) {
		d.push_back(s[i] - '0');
	}
	return d;
}
map< ll, int > prime_factor(ll n) {
	map< ll, int > ret;
	for (ll i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			ret[i]++;
			n /= i;
		}
	}
	if (n != 1) ret[n] = 1;
	return ret;
}

vector<ll> Enumeration(ll n) {
	vector<ll>res;
	for (ll i = 1; i * i <= n; i++) {
		if (n % i != 0)continue;
		res.push_back(i);
		if (i * i != n)res.push_back(n / i);
	}
	
	return res;
}



ll myPow(ll x, ll n, ll m) {//x^n
	if (n == 0)
		return 1;
	if (n % 2 == 0)
		return myPow(x * x % m, n / 2, m);
	else
		return x * myPow(x, n - 1, m) % m;
}
std::vector<int> Eratosthenes(const int N)//素数列挙
{
	std::vector<bool> is_prime(N + 1);
	for (int i = 0; i <= N; i++)
	{
		is_prime[i] = true;
	}
	std::vector<int> P;
	for (int i = 2; i <= N; i++)
	{
		if (is_prime[i])
		{
			for (int j = 2 * i; j <= N; j += i)
			{
				is_prime[j] = false;
			}
			P.emplace_back(i);
		}
	}
	return P;
}
const ll mod = MOD;

struct mint {

	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x% mod + mod) % mod) {}
	
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this; }
	mint operator+(const mint a) const { return mint(*this) += a; }
	mint operator-(const mint a) const { return mint(*this) -= a; }
	mint operator*(const mint a) const { return mint(*this) *= a; }
	
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime mod
	mint inv() const { return pow(mod - 2); }
	mint& operator/=(const mint a) { return *this *= a.inv(); }
	mint operator/(const mint a) const { return mint(*this) /= a; }
	
	//operator long long() const { return x; }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
struct combination {//comb(a,b)でよぶaCb

	vector<mint> fact, ifact;
	combination(ll n) :fact(n + 1), ifact(n + 1) {
		assert(n < mod);
		fact[0] = 1;
		for (ll i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (ll i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(ll n, ll k) {
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
}comb(1000005);//comb貯め
mint f(ll n, ll k) {
	if (n < 0) return 0;
	// nPk = nCk * k!
	mint res = comb(n, k);
	res *= comb.fact[k];
	return res;
}




struct SegmentTree {
private:
	int n;
	vector<int> node;

public:
	SegmentTree(vector<int> v) {//SegmentTree seg(vector<int>(N,INF))で呼び出し 
		int sz = v.size();
		n = 1; while (n < sz) n *= 2;
		node.resize(2 * n - 1, INF);
		for (int i = 0; i < sz; i++) node[i + n - 1] = v[i];
		for (int i = n - 2; i >= 0; i--) node[i] = min(node[2 * i + 1], node[2 * i + 2]);
	}

	void update(int x, int val) {
		x += (n - 1);
		node[x] = val;
		while (x > 0) {
			x = (x - 1) / 2;
			node[x] = min(node[2 * x + 1], node[2 * x + 2]);
		}
	}

	int getmin(int a, int b, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = n;
		if (r <= a || b <= l) return INF;
		if (a <= l && r <= b) return node[k];

		int vl = getmin(a, b, 2 * k + 1, l, (l + r) / 2);
		int vr = getmin(a, b, 2 * k + 2, (l + r) / 2, r);
		return min(vl, vr);
	}
};
// 1-indexedなので注意。
struct BIT {
private:
	vector<int> bit;
	int N;
public:
	BIT(int size) {
		N = size;
		bit.resize(N + 1);
	}

	// 一点更新です
	void add(int a, int w) {//a番目をwに更新
		for (int x = a; x <= N; x += x & -x) bit[x] += w;
	}

	// 1~Nまでの和を求める。
	int sum(int a) {
		int ret = 0;
		for (int x = a; x > 0; x -= x & -x) ret += bit[x];
		return ret;
	}
};
/****Union-Ford-Tree***/
/*int par[int(0x7ffffff)];//Parent
int Rank[int(0x7ffffff)];//Deep_of_the_Tree

//int a[100] = { 4,2,3,1,5 }, m[100] ={3,2,2};
//int n = 0;
//const ll mod = 1000000007;


//n要素で初期化
void init(int n) {
	for (int i = 0; i < n; i++) {
		par[i] = i;
		Rank[i] = 0;
	}
}
//木の根探し
int find(int x) {
	if (par[x] == x) {
		return x;
	}
	else {
		return par[x] = find(par[x]);
	}
}
//xとyの属する集合を併合
void unite(int x, int y) {
	x = find(x);
	y = find(y);
	if (x == y)return;

	if (Rank[x] < Rank[y]) {
		par[x] = y;
	}
	else {
		par[y] = x;
		if (Rank[x] == Rank[y])Rank[x]++;
	}
}

bool same(int x, int y) {
	return find(x) == find(y);
}*/
//queue古い順,stack新しい順
/************************************************************************************/
int N, T; int A[5000], B[5000]; int dp[10000][10000];
int32_t main() {
	cin >> N >> T;
	rep(i, N) {
		cin >> A[i] >> B[i];
	}
	int ans = 0;
	rep(i, N) {
		rep(j, T+3500) {
			if (j < A[i])dp[i + 1][j] = dp[i][j];
			else dp[i + 1][j] = max(dp[i][j], dp[i][j-A[i]] + B[i]);
			if (j - A[i] < T&&j-A[i]>=0) {
				chmax(ans, dp[i][j-A[i]]+B[i]);
			}
			else if (j < T) {
				chmax(ans, dp[i + 1][j]);
			}

		}
	}
	cout << ans << endl;

	
}