Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.

But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!

3

10 110

2

1 1

30 50

10 110

2

1 1

50 30

1 6

2

10 3

20 4

The minimum amount of money in the piggy-bank is 60.

The minimum amount of money in the piggy-bank is 100.

This is impossible.

// Asimple

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <cstdlib>

#include <queue>

#include <vector>

#include <string>

#include <cstring>

#include <stack>

#include <set>

#include <map>

#define INF 0x3f3f3f3f

#define mod 1000000007

#define debug(a) cout<<#a<<" = "<<a<<endl

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int maxn = +;

int n, m, T, len, cnt, num, ans, Max;

int x, y;

int v[maxn], w[maxn];

int dp[maxn];

void input() {

    //cin >> T;

    scanf("%d", &T);

    while( T -- ) {

        //cin >> x >> y;

        scanf("%d%d", &x, &y);

        m = y - x;

        //memset(dp, INF, sizeof(dp));

        for(int i=; i<=m; i++) dp[i] = INF;

        dp[] = ;

        //cin >> n;

        scanf("%d", &n);

        for(int i=; i<=n; i++) //cin >> v[i] >> w[i];

            scanf("%d%d", &v[i], &w[i]);

        for(int i=; i<=n; i++) {

            for(int j=w[i]; j<=m; j++) {

                dp[j] = min(dp[j], dp[j-w[i]]+v[i]);

            }

        }

        if( dp[m]==INF ) printf("This is impossible.\n");

        else printf("The minimum amount of money in the piggy-bank is %d.\n", dp[m]);

        //if( dp[m]==INF ) cout << "This is impossible." << endl;

        //else cout << "The minimum amount of money in the piggy-bank is " << dp[m] << "." << endl;

    }

}

int main() {

    input();

    return ;

}