#include<bits/stdc++.h>
using namespace std;
#define endl "\n"
#define debug(x) cout<<#x<<": "<<x<<endl
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
using ll=long long;
using ull=unsigned long long;
void solve(){
    int n,f;
    cin>>n>>f;
    vector<int>a(n),b(n);
    for(int i = 0;i<n;i++)
        cin>>a[i]>>b[i];
    vector<int>ord(n);
    iota(ord.begin(),ord.end(),0);
    sort(ord.begin(),ord.end(),[&](const int &i,const int &j){
        if(b[i] == b[j])
            return a[i] < a[j];
        return b[i] < b[j];
    });
    int d = 0;
    for(;d < n;d++)
    {
        if(b[ord[d]] >= f)
            break;
    }
    for(;d - 1 >= 0;d--)
    {
        if(b[ord[d - 1]] < a[ord[d]])
            break;
    }
    ll ans = 0;
    int lst = f;
    for(int k = d;k<n;k++)
    {
        int i = ord[k];
        ans += (max(a[i] - lst,0) + b[i] - a[i]);
        lst = b[i];
    }
    for(int k = 0;k < d;k++)
    {
        int i = ord[k];
        ans += (max(a[i] - lst,0) + b[i] - a[i]);
        lst = b[i];
    }
    cout<<ans<<endl;
    for(int k = d;k<n;k++)
        cout<<ord[k]+1<<" ";
    for(int k = 0;k < d;k++)
        cout<<ord[k]+1<<" ";
    cout<<endl;
    
}
signed main(){
    ios::sync_with_stdio(0);
    cout.tie(0);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--){
        solve();
    }
}
/*
贡献法
正难则反
数小状压
关系连边
拆位
广义单调性
最长转最短
*/