// 考虑绝对值很难做，分别做大于和小于的部分。
// 对于小于，有贡献的形如 ti < tj 且 ai < hj. 这是一个经典(?)的双指针cdq

#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i = a; i <= b; ++i)
using namespace std;

inline int rd() {
    int r = 0; bool w = false; char ch = getchar();
    while( ch < '0' || ch > '9' ) w = !(ch^45), ch = getchar();
    while( ch >= '0' && ch <= '9' ) r = (r<<1) + (r<<3) + (ch^48), ch = getchar();
    return w ? -r : r;
}

#define MAXM 1000000
#define MAXN 500000
int n, m;
int x[MAXM+5], h[MAXM+5];
int L[MAXM+5], R[MAXM+5];
int ans[MAXM+5];

int tots, tote;
struct Soldier {
    int x, h;
    Soldier() {}
    Soldier(int xi, int hi): x(xi), h(hi) {}
}bucs[MAXM+5];
struct Enemy {
    int l, r, h, idx;
    Enemy() {}
    Enemy(int li, int ri, int hi, int i): l(li), r(ri), h(hi), idx(i) {}
}buce[MAXM+5];

int a[MAXN*4+5];
#define lson (p<<1)
#define rson ((p<<1)+1)
const int INF = 1e9;
void pushup( int p, int typ ) {
    if( a[lson] == -1 || a[rson] == -1 ) a[p] = ((a[lson]+1) | (a[rson]+1)) - 1; 
    else a[p] = typ ? max(a[lson], a[rson]) : min(a[lson], a[rson]);
}
void update( int p , int x , int k , int typ , int L = 1, int R = n ) {
    if( L == R ) {
        a[p] = k;
        return;
    }
    int mid = (L + R) >> 1;
    if( x <= mid ) update(lson, x, k, typ, L, mid);
    else update(rson, x, k, typ, mid+1, R);
    pushup(p, typ);
}
void refresh( int p , int x , int typ , int L = 1 , int R = n ) {
    if( L == R ) {
        a[p] = -1;
        return;
    }
    int mid = (L + R) >> 1;
    if( x <= mid ) refresh(lson, x, typ, L, mid);
    else refresh(rson, x, typ, mid+1, R);
    pushup(p, typ);
}
int query( int p , int l , int r , int typ , int L = 1 , int R = n ) {
    if( l == L && r == R  ) {
        return a[p];
    }
    int mid = (L + R) >> 1;
    if( r <= mid ) return query(lson, l, r, typ, L, mid);
    else if( l > mid ) return query(rson, l, r, typ, mid+1, R);
    else {
        int ql = query(lson, l, mid, typ, L, mid);
        int qr = query(rson, mid+1, r, typ, mid+1, R);
        if( ql == -1 || qr == -1 ) return ((ql + 1) | (qr + 1)) - 1;
        return typ ? max(ql, qr) : min(ql, qr);
    }
}

void cdq( int l , int r ) {
    if( l == r ) return;
    int mid = (l + r) >> 1;
    
    tots = tote = 0;
    FOR(i,l,mid) if( x[i] != 0 ) {
        bucs[++tots] = Soldier(x[i], h[i]);
    }
    FOR(i,mid+1,r) if( L[i] != 0 ) {
        buce[++tote] = Enemy(L[i], R[i], h[i], i);
    }

    sort(bucs+1, bucs+1+tots, [](const Soldier &a, const Soldier &b) {
        return a.h < b.h;
    });
    sort(buce+1, buce+1+tote, [](const Enemy &a, const Enemy&b) {
        return a.h < b.h;
    });
    int p = 0;
    FOR(i,1,tote) {
        while( p < tots && bucs[p+1].h <= buce[i].h ) {
            ++p;
            update(1, bucs[p].x, bucs[p].h, 1);
        }
        int cons = query(1, buce[i].l, buce[i].r, 1);
        if( cons != -1 ) {
            if( ans[buce[i].idx] == -1 ) ans[buce[i].idx] = buce[i].h - cons;
            else ans[buce[i].idx] = min(ans[buce[i].idx], buce[i].h - cons);
        }
    }
    FOR(i,1,p) refresh(1, bucs[i].x, 1);

    sort(bucs+1, bucs+1+tots, [](const Soldier &a, const Soldier &b) {
        return a.h > b.h;
    });
    sort(buce+1, buce+1+tote, [](const Enemy &a, const Enemy&b) {
        return a.h > b.h;
    });
    p = 0;
    FOR(i,1,tote) {
        while( p < tots && bucs[p+1].h > buce[i].h ) {
            ++p;
            update(1, bucs[p].x, bucs[p].h, 0);
        }
        int cons = query(1, buce[i].l, buce[i].r, 0);
        if( cons != -1 ) {
            if( ans[buce[i].idx] == -1 ) ans[buce[i].idx] = cons - buce[i].h;
            else ans[buce[i].idx] = min(ans[buce[i].idx], cons - buce[i].h);
        }
    }
    FOR(i,1,p) refresh(1, bucs[i].x, 0);

    cdq(l, mid);
    cdq(mid+1, r);
}
void solve() {
    n = rd(), m = rd();
    FOR(i,1,m) ans[i] = -1;
    FOR(i,1,m) L[i] = R[i] = x[i] = h[i] = 0;
    FOR(i,1,m) {
        int typ = rd();
        if( typ == 0 ) {
            x[i] = rd(), h[i] = rd();
        } else {
            L[i] = rd(), R[i] = rd(), h[i] = rd();
        }
    }
    cdq(1, m);
    FOR(i,1,m) if( L[i] ) printf("%d\n", ans[i]);
}

int main() {
    memset(a, -1, sizeof(a));

    int T = rd(); while( T-- ) solve();    
    return 0;
}