Problem:
Score : 100 points

Problem Statement

We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand.

How many grams of sand will the upper bulb contains after t seconds?

Constraints

1≤X≤10^9

1≤t≤10^9

X and t are integers.

Input

The input is given from Standard Input in the following format:

X t

Output

Print the number of sand in the upper bulb after t second.

Sample Input 1

100 17

Sample Output 1

83

17 out of the initial 100 grams of sand will be consumed, resulting in 83 grams.

Sample Input 2

48 58

Sample Output 2

0

All 48 grams of sand will be gone, resulting in 0 grams.

Sample Input 3

1000000000 1000000000

Sample Output 3

0

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

#include <vector>
#include <math.h>
#include <iostream>
#include <string>
#include <algorithm>
#include <queue>
#include <iomanip>
#include <deque>

using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
template <class T> using V = vector<T>;
template <class T> using VV = V<V<T>>;
#define rep(i, n) for(int i = 0; i<n; i++)
#define rep_s(i, start, n) for(int i= start;  i<n; i++)
#define ALL(a) (a).begin(), (a).end()
string alphabet_upper = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string alphabet_lower  = "abcdefghijklmnopqrstuvwxyz";

/*
using edge = struct {int to; ll cost;};
vector<edge> tree[limit];
ll depth[limit];

void dfs(int v, int p, ll d){
  depth[v] = d;
  //cout <<"v: " << v << endl;
  for (auto &e: tree[v]){
    //cout << "e.to: " << e.to << endl;
    if (e.to == p) continue;
    dfs(e.to, v, d+e.cost);
  }
}
*/

int main(void){
  int X, t;
  cin >> X >> t;
  if (X-t > 0) cout << X-t << endl;
  else cout << 0 << endl;
}