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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

function main(input){
  const lines = input.split('\n');
  const NW = lines.shift().split(' ').map(x => parseInt(x));
  const N = NW[0]
  const W = NW[1]
  const items = lines.map(line => line.split(' ').map(x => parseInt(x)));


  const first = items.shift();
  const first_weight = first[0];
  const first_value = first[1];
  var value_sum = first_value;
  dp = [0]

  for(var i=1;i<=value_sum;i++){
    dp[i]=first_weight;
  }



  for(var i=0;i<items.length;i++){
    var current_weight = items[i][0];
    var current_value = items[i][1];
    value_sum += current_value;
  //console.log(dp);
    for(var value=value_sum;value>0;value--){



      if(value >= current_value){
        var prev_weight = dp[value - current_value];

        if(dp[value]==undefined){
          dp[value] = current_weight + prev_weight;

        }else{
          dp[value] = Math.min(prev_weight+current_weight, dp[value]);

        }
      }else{

        if(dp[value]==undefined){
          dp[value]=current_weight;
        }else{
          dp[value]=Math.min(current_weight, dp[value]);

        }


     }

//valueがcurrent_value以上なら、合算weightを入れられる
//合算weightはその前のマスにある値(weight)とcurrent_weightを足したもの

//その時、今のマス目を見みて入れ方を変える
//今のマス目インデックス(value)に何もなければcurrent_weightをそのまま入れる、
// 今のマス目に何かあれば、前のマス目の値(weight)とcurrent_weightを足したものとで値(weight)を比較、低い方を入れる。


// current_valueがvalue未満ならcurrent_weightを入れるが、マス目に値があるかどうかで分岐する。
// マス目に何も入っていなければcurrent_weightを入れる
// なにか入っていれば比較を行い、軽い方を入れる。


//そのインデックスになにかある、かつcurrent_valueがvalueよりも大きいとそのvalueはそこに入らないので、そのままにしておく
// 

    }

  }


  var result = 0;

  for(var i=value_sum;i>=0;i--){
    if(dp[i]<W){
      result = i;
      break;
    }
  }

  //console.log(dp);

  console.log(result);

}

main(require('fs').readFileSync('/dev/stdin', 'utf8'));