We choose $c$ between $a$ and $b$ $(a≤c≤b)$. The distance is $(c−a)+(b−c)=b−a$. Note that the distance does not depend on the the position $c$ at all.

/**
 *    author: 小飞侠
 *    created: 2024.11.06 23:15:40
 */
#include <bits/stdc++.h>
using namespace std;
#define ls u << 1
#define rs u << 1 | 1
#define LL long long
#define int long long
#define PII pair <int, int>
#define fi first
#define se second
#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define lb lower_bound
#define ub upper_bound
#define i128 __int128
#define pcnt(x) __builtin_popcount(x)
#define mem(a,goal) memset(a, (goal), sizeof(a))
#define rep(x,start,end) for(int x = (start) - ((start) > (end)); x != (end) - ((start) > (end)); ((start) < (end) ? x ++ : x --))
#define aLL(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
const int INF = 998244353;
const int P = 1e9 + 7;
const int N = 2e5 + 5;
void solve()
{
    int a, b;
    cin >> a >> b;
    cout << b - a << '\n';
}
signed main()
{
    int t;
    cin >> t;
    while(t --) solve();
    return 0;
}