Problem:
Score : 100 points

Problem Statement

There are N items, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i.

Taro has decided to choose some of the N items and carry them home in a knapsack.
The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W.

Find the maximum possible sum of the values of items that Taro takes home.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq W \leq 10^9

1 \leq w_i \leq W

1 \leq v_i \leq 10^3

Input

Input is given from Standard Input in the following format:

N W
w_1 v_1
w_2 v_2
:
w_N v_N

Output

Print the maximum possible sum of the values of items that Taro takes home.

Sample Input 1

3 8
3 30
4 50
5 60

Sample Output 1

90

Items 1 and 3 should be taken.
Then, the sum of the weights is 3 + 5 = 8, and the sum of the values is 30 + 60 = 90.

Sample Input 2

1 1000000000
1000000000 10

Sample Output 2

10

Sample Input 3

6 15
6 5
5 6
6 4
6 6
3 5
7 2

Sample Output 3

17

Items 2, 4 and 5 should be taken.
Then, the sum of the weights is 5 + 6 + 3 = 14, and the sum of the values is 6 + 6 + 5 = 17.

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

#include <bits/stdc++.h>
using namespace std;
/*#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
// x^y≡x^(y%(p-1))mod p
// but there is a condition that x needs to be coprime to p
///assic value of ('0'-'9') is(48 - 57) and (a-z) is (97-122) and (A-Z) is(65-90) and 32 for space
typedef tree<int, null_type, less<int>, rb_tree_tag,
             tree_order_statistics_node_update>
    indexed_set; */
typedef long long int ll;
typedef pair<ll, ll> pii;
typedef vector<pii> vii;
#define  pb push_back
#define f(i,a,b) for(ll i=a;i<b;i++)
#define fo(i,a,b) for(ll i = a; i<=b;i+=1)
#define rf(i,a,b)  for(ll i=a;i>=b;i--)
#define vll vector<ll>
#define tests() int test_cases ; cin >> test_cases ; while(test_cases--)
#define ub upper_bound
#define lb lower_bound
#define all(v) v.begin(),v.end()
#define MAXN 100010
#define MOD 1000000007
#define mod 998244353
#define deb(x) cout << '>' << #x << ':' << x << endl;
#define sp " "
#define MS(x,y) fill_n(*x, sizeof x / sizeof **x, y);
#define mem(x,y) memset(x,y,sizeof(x));
#define INF 1000000000000000000
#define IO                 \
	ios::sync_with_stdio(0); \
	cin.tie(0);              \
	cout.tie(0);
#define debug(x) cerr << (#x) << " is " << (x) << endl;

#ifndef ONLINE_JUDGE
#define cerr if(true)cerr
#endif

template<int D, typename T>
struct Vec : public vector<Vec<D - 1, T>> {
static_assert(D >= 1, "Vector dimension must be greater than zero!");
template<typename... Args>
	Vec(int n = 0, Args... args) : vector<Vec<D - 1, T>>(n, Vec<D - 1, T>(args...)) {
	}
};
template<typename T>
struct Vec<1, T> : public vector<T> {
	Vec(int n = 0, const T& val = T()) : vector<T>(n, val) {
	}
};

ll power(ll x, ll y,ll p)
{
    ll res = 1;     // Initialize result
    x = x % p;
    if (y==0)return 1;
    if(x==0)return 0;
    while (y > 0)
    {
        if (y & 1)
            res = (res*x) % p;
        y = y>>1; // y = y/2
        x = (x*x) % p;  // Change x to x^2
    }
    return res;
}
int gcd(int a, int b) { if (a == 0)return b;return gcd(b % a, a);}

ll modinv(ll a,ll p){return power(a,p-2,p);}


void solve(){
    ll n,W;
    cin>>n>>W;
    Vec<1,ll> w(n),v(n);
    f(i,0,n){
        cin>>w[i]>>v[i];
    }
    Vec<2,ll> dp(n,MAXN,INF);
    dp[0][v[0]] = w[0];
    dp[0][0] = 0;
    fo(i,1,n-1){
        f(j,0,MAXN){
            dp[i][j] = dp[i-1][j];
            if(j-v[i]>=0)
            dp[i][j] = min(dp[i][j], dp[i-1][j-v[i]] +w[i]);
        }
    }
    ll mx = 0;
    f(j,0,MAXN){
        if (dp[n-1][j]<=W)
        {
            mx = max(mx, j);
        }
    }
    cout<<mx<<endl;
    return ;
}
int main()

{
    IO;
    int T=1;
	// cin >> T;
	f(c,0,T){
		solve();
	}
    return 0;
}// make_pair make_tuple  tuple<int,int,int> continue resize insert count size else true isValid second first false erase