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
#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <queue>
#include <map>
using namespace std;
typedef long long ll;
const int INF = 1<<30; 
const int MOD = 1e9 + 7;
const int dy[] = {1, 0, -1, 0};
const int dx[] = {0, 1, 0, -1};
const long long MAX_V = 1000 * 110;
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, w; cin >> n >> w;
    int val[n + 1], weight[n + 1];
    for(int i = 0; i < n; i++)  cin >> val[i] >> weight[i];
    vector<vector<int>> dp(n + 1, vector<int>(MAX_V + 1, INF));
    dp[0][0] = 0;
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j <= MAX_V; j++)
        {
            if(j - val[i] < 0)
            {
                dp[i + 1][j] = dp[i][j];
            }
            else 
            {
                dp[i + 1][j] = min(dp[i][j], dp[i][j - val[i]] + weight[i]);
            }
        }
    }
    int ans;
    for(int v = 0; v <= MAX_V; v++)
    {
        if(dp[n][v] <= w) ans = v;
    }
    cout << ans << endl;
    return 0;
}


