#include<bits/stdc++.h>

using namespace std;

#define ll int64_t
#define pii pair<int ,int >
#define pllll pair<ll ,ll >
#define p(x,y) pair<x,y>
#define mkp make_pair

//得到坐标,但是并不是矩形坐标，而是数字三角形的坐标
/*
pllll getcoord(ll z)
{
    ll x=0;
    //这个写法是什么，倍增倍减吗？
    //二进制优化，基的概念去表示一个数字
    for(int i=1<<30;i;i>>=1){
        if( (x|i)*((x|i)+1)/2 <= z )
            x|=i;
    }
    //x是指行数
    return {x,z-x*(x+1)/2};
}

void solve()
{
    int64_t x, y;
    cin >> x >> y;
    if (x > y) {
        cout << x - y << '\n';
        return;
    }

    auto [sx, sy] = getCoord(x - 1);
    auto [tx, ty] = getCoord(y - 1);
    int64_t ans = 1e18;

    debug(sx, sy);
    debug(tx, ty);
    assert (sx <= tx);
    if (tx - ty >= sx - sy) {
        ans = min(ans, tx - sx + ((tx - ty) - (sx - sy)));
    }
    {
        ans = min(ans, tx - sx + 1 + (tx - ty) + (sy + 1));
    }

    cout << ans << '\n';
}


*/


ll mysqrt(ll x){ll s=max(0ll,(ll)sqrt(x)-3ll);while( (s+1)*(s+1)<=x )++s;return s;}
ll upstep(ll x){return (mysqrt(x<<3)+3)/2;}
ll mmul(ll a,ll b){if ((double)a * b > 6e18)return 6e18;else return a*b;}
ll jump(ll x,ll cnt){ll s=upstep(x);return x+mmul(cnt,s+s+cnt-1)/2;}
ll  prev_crit(ll x)
{
    ll n=mysqrt(x<<1)+5;
    while(n*(n+1)/2+1 > x)--n;
    return n*(n+1)/2+1;
}

signed main(int argc,char** argv)
{
    ios::sync_with_stdio(false);

    int T;
    cin>>T;
    while(T--)
    {
        ll x,y;
        cin>>x>>y;
        if(x>=y){
            cout<<x-y<<'\n';
            continue;;
        }
        ll r=1;
        while(jump(x,r) < y)r<<=1;
        ll l=0;
        while(r-l>1)
        {
            ll mid=l+r>>1;
            if( jump(x,mid) < y )l=mid;
            else r=mid;
        }

        ll ss=r;
        ll ans=0;
        while(1)
        {
            //目标和左端点相差多远
            ll diff=jump(x,ss)-y;
            //压根没差，直接中奖
            if(!diff)break;
            //左端点的数值pre
            ll pre=prev_crit(x);
            if(diff<=x-pre)
            {//如果还能待在这里层数
                x-=diff;
                ans+=diff;
                break;
            }
            ans+=x-pre;
            //x调整为左端点，继续跳步
            x=pre;
            //目标和左端点相差多远
            diff = jump(x,ss)-y;
            //压根没差，直接中奖
            if(!diff)break;
            //如果x跳步后和目标多出来的的数值还比跳的次数多
            //那就选择减小，这样更优
            if(ss+1<=diff)
            {
                --x;
                ++ans;
                continue;
            }
            //否则如果贸然减小的话造成的影响会
            ans+=diff;
            break;
        }
        cout<<ans+ss<<'\n';
    }
    return 0;
}