Problem: You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.

You want to find a subset of items to put such that:

* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.



Find the maximum total value of items in the knapsack.

Constraints

* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000

Input


N W
v1 w1
v2 w2
:
vN wN


The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.

Output

Print the maximum total values of the items in a line.

Examples

Input

4 5
4 2
5 2
2 1
8 3


Output

13


Input

2 20
5 9
4 10


Output

9

Solution (in CPP):

Solution:
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
// #define USE_LLONG_AS_INT
#ifdef USE_LLONG_AS_INT
#define int long long
#define inf (1ll<<60)
#else
#define inf (1<<30)
#endif
#define rep(i,n) for(int i=0;i<n;i++)
#define Rep(i,a,b) for(int i=a;i<b;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define rev(i,n) for(int i=(n)-1;i>=0;i--)
#define vi vector<int>
#define vvi vector<vi>
#define pb push_back
#define pi pair<int,int>
#define vp vector<pair<int,int>>
#define mp make_pair
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
#define MEMSET(a) memset(a,0,sizeof(a))
#define Yes(f) cout<<(f?"Yes":"No")<<endl
#define yes(f) cout<<(f?"yes":"no")<<endl
#define YES(f) cout<<(f?"YES":"NO")<<endl
#define SORT(v) sort(all(v))
#define RSORT(v) sort(all(v), greater<int>())

using namespace std;

const int mod=1e9+7;
const string sp=" ";

void run();

void init() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout<<fixed<<setprecision(12);
}

signed main(){
    init();
    run();
    return 0;
}

#define N 100
#define V 100

int dp[N+1][N*V+1];

void run(){
    int n,W;
    cin>>n>>W;
    vi v(n),w(n);
    rep(i,n)cin>>v[i]>>w[i];
    fill(dp[0],dp[0]+N*V+1,inf);
    dp[0][0]=0;
    rep(i,n){
        rep(j,N*V+1){
            if(j<v[i])dp[i+1][j]=dp[i][j];
            else dp[i+1][j]=min(dp[i][j],dp[i][j-v[i]]+w[i]);
        }
    }
    int ans=0;
    rep(i,N*V+1)if(dp[n][i]<=W)ans=i;
    cout<<ans<<endl;
}