#include <bits/stdc++.h>
using namespace std;

#define For(i, a, b) for(int i = (a); i < (b); i++)
#define rep(i, n) For(i, 0, n)
#define rFor(i, a, b) for(int i = (a); i >= (b); i--)
#define ALL(v) (v).begin(), (v).end()
#define rALL(v) (v).rbegin(), (v).rend()
#define SZ(v) ((int)(v).size())

using lint = long long;
using ld = long double;

const int INF = 2000000000;
const lint LINF = 1000000000000000000;
// 真上から反時計回り
const int di[] = {-1, 0, 1, 0};
const int dj[] = {0, -1, 0, 1};
const int di8[] = {-1, -1, 0, 1, 1, 1, 0, -1};
const int dj8[] = {0, -1, -1, -1, 0, 1, 1, 1};

struct SetupIo {
    SetupIo() {
        ios::sync_with_stdio(false); cin.tie(nullptr);
        cout << fixed << setprecision(15);
        cerr << fixed << setprecision(15);
    }
} setupio;

template <class T>
bool chmin(T &a, const T &b) {
    return a > b ? a = b, 1 : 0;
}
template <class T>
bool chmax(T &a, const T &b) {
    return a < b ? a = b, 1 : 0; 
}
template <class T>
T mypow(T a, T b) {
    T res = 1;
    while (b) {
        if (b & 1) { res *= a; }
        a *= a; b >>= 1;
    }
    return res;
}
template <class T>
T modpow(T a, T b, T mod) {
    T res = 1;
    while (b) {
        if (b & 1) { res = (res * a) % mod; }
        a = (a * a) % mod; b >>= 1;
    }
    return res;
}

int main() {
    lint N, P, Q;
    cin >> N >> P >> Q;
    vector<lint> X(N), A(N), B(N);
    rep(i, N) {
        cin >> X[i] >> A[i] >> B[i];
    }

    // 自明な No その 1
    rep(i, N) {
        if (A[i] + B[i] < X[i]) {
            cout << "No\n";
            return 0;
        }
    }

    // 自明な No その 2
    if (accumulate(ALL(X), 0LL) > P + Q) {
        cout << "No\n";
        return 0;
    }
    
    // X[i] - B[i] <= a[i] <= A[i]
    // sum_X - Q <= sum_a <= P
    vector<lint> a(N);

    // 一旦最低限割り当てる
    rep(i, N) {
        a[i] = X[i] - B[i];
    }

    lint sum_a = accumulate(ALL(a), 0LL);

    // 2 つの条件を満たしながら a[i] を増やしていく
    rep(i, N) {
        lint add = min(A[i] - a[i], P - sum_a);
        a[i] += add;
        sum_a += add;
    }

    if (accumulate(ALL(X), 0LL) - Q <= sum_a) {
        cout << "Yes\n";
        rep(i, N) {
            cout << a[i] << " " << X[i] - a[i] << "\n";
        }
    } else {
        cout << "No\n";
    }
}